CHAPTER 2 1* What is the approximate average velocity of the race cars during the Indianapolis 500? Since the cars go around a closed circuit and return nearly to the starting point, the displacement is nearly zero, and the average velocity is zero. 2 Does the following statement make sense? The average velocity of the car at 9 a.m. was 60 km/h. No, it does not. Average velocity must refer to a finite time interval. 3 Is it possible for the average velocity of an object to be zero during some interval even though its average velocity for the first half of the interval is not zero? Explain. Yes, it is. In a round trip, A to B and back to A, the average velocity is zero; the average velocity between A and B is not zero. 4 The diagram in Figure 2-21 tracks the path of an object moving in a straight line. At which point is the object farthest from its starting point? (a) A (b) B (c) C (d) D (e) E (b) Starting point is at x = 0; point B is farthest from x = 0. 5* (a) An electron in a television tube travels the 16-cm distance from the grid to the screen at an average speed of 4 10 7 m/s. How long does the trip take? (b) An electron in a current-carrying wire travels at an average speed of 4 10-5 m/s. How long does it take to travel 16 cm? (a) From Equ. 2-3, t = s/(av. speed) (b) Repeat as in (a) t = (0.16 m)/(4 10 7 m/s) = 4 10-9 s = 4 ns t = (0.16 m)/(4 10-5 m/s) = 4 10 3 s = 4 ks 6 A runner runs 2.5 km in 9 min and then takes 30 min to walk back to the starting point. (a) What is the runner s average velocity for the first 9 min? (b) What is the average velocity for the time spent walking? (c) What is the average velocity for the whole trip? (d) What is the average speed for the whole trip? Take the direction of running as the positive direction. (a) Use Equ. 2-2 (b) Use Equ. 2-2 (c) x = 0 (d) Total distance = 5.0 km; t = 39 min v av = (2.5 km)/[(9 min)(1 h/60 min)] = 16.7 km/h v av = (-2.5 km)/(0.5 h) = -5.0 km/h v av = 0 Av. speed = (5.0 km)/[(39 min)(1 h/60 min)] = 7.7 km/h
7 A car travels in a straight line with an average velocity of 80 km/h for 2.5 h and then with an average velocity of 40 km/h for 1.5 h. (a) What is the total displacement for the 4-h trip? (b) What is the average velocity for the total trip? (a) 1. Find the displacements of each segment; Use Equ. 2-2 2. Add the two displacements (b) Use Equ. 2-2 S 1 = (80 km/h)(2.5 h) = 200 km; S 2 = (40 km/h)(1.5 h) = 60 km S tot = 260 km v av = (260 km)/(4 h) = 65 km/h 8 One busy air route across the Atlantic Ocean is about 5500 km. (a) How long does it take for a supersonic jet flying at 2.4 times the speed of sound to make the trip? Use 340 m/s for the speed of sound. (b) How long does it take a subsonic jet flying at 0.9 times the speed of sound to make the same trip? (c) Allowing 2 h at each end of the trip for ground travel, check-in, and baggage handling, what is your average speed door to door when traveling on the supersonic jet? (d) What is your average speed taking the subsonic jet? (a) Find v, then use Equ. 2-2 v = 2 (0.340 km/s)(3600 s/h) = 2448 km/h; t = (5500 km)/(2448 km/h) = 2.25 h (b) Repeat as for part (a) v = 0.9 (0.340 km/s)(3600 s/h) = 1102 km/h; t = (5500 km)/(1102 km/h) = 5.0 h (c) Find the total time, then use Equ. 2-2 t tot = (2.25 + 4)h = 6.25 h; v av = (5500 km)/(6.25 h) = 880 km/h (d) Repeat as for part (c) t tot = 9 h; v av = (5500 km)/(9 h) = 611 km/h 9* As you drive down a desert highway at night, an alien spacecraft passes overhead, causing malfunctions in your speedometer, wristwatch, and short-term memory. When you return to your senses, you can t tell where you are, where you are going, or even how fast you are traveling. The passenger sleeping next to you never woke up during this incident. Although your pulse is racing, hers is steady at 55 beats per minute. (a) If she has 45 beats between the mile markers posted along the road, determine your speed. (b) If you want to travel at 120 km/h, how many heartbeats should there be between mile markers? (a) Find the time between mile markers t = (45 beats/mile)/(55 beats/min) = 0.818 min v = s/ t v = 1 mi/0.818 min = 1.22 mi/min = 73.3 mi/h (b) N = (beats/min)(60 min/h)/(v mi/h) N = (55 60 beats/h)/[(120 km/h)/(1.61 km/mi)] = 44.3 10 The speed of light, c, is 3 10 8 m/s. (a) How long does it take for light to travel from the sun to the earth, a distance of 1.5 10 11 m? (b) How long does it take light to travel from the moon to the earth, a distance of 3.84 10 8 m? (c) A light-year is a unit of distance equal to that traveled by light in 1 year. Convert 1 light-year into kilometers and miles. (a) Use Equ. 2-3 t = (1.5 10 11 m)/(3 10 8 m/s) = 500 s (b) Use Equ. 2-3 t = (3.84 10 8 m)/(3 10 8 m/s) = 1.28 s (c) See Problem 1-54(c) 1 l-y = 9.48 10 15 m = 9.48 10 12 km Express km in miles = (9.48 10 12 km)(1 mi/1.61 km) = 5.9 10 12 mi.
11 The nearest star, Proxima Centauri, is 4.1 10 15 km away. From the vicinity of this star, Gregor places an order at Tony s Pizza in Hoboken, New Jersey, communicating via light signals. Tony s fastest delivery craft travels at 10-4 c (see Problem 10). (a) How long does it take Gregor s order to reach Tony s Pizza? (b) How long does Gregor wait between sending the signal and receiving the pizza? If Tony s has a 1000-years-or-it s-free delivery policy, does Gregor have to pay for the pizza? (a) Use Equ. 2-3 t = (4.1 10 18 m)/(3 10 8 m/s) = 1.37 10 10 s = 434 y (b) Traveling at 10-4 c, time will be 10 4 times (a) t = 4.34 10 6 y; he need not pay for the pizza 12 A car making a 100-km journey travels 40 km/h for the first 50 km. How fast must it go during the second 50 km to average 50 km/h? 1. Find the time for the total journey t tot = (100 km)/(50 km/h) = 2 h 2. Find the time for the first 50 km t 1 = (50 km)/(40 km/h) = 1.25 h 3. Find the speed in the remaining 0.75 h v = (50 km)/(0.75 h) = 66.7 km/h 13* John can run 6.0 m/s. Marcia can run 15% faster than John. (a) By what distance does Marcia beat John in a 100-m race? (b) By what time does Marcia beat John in a 100-m race? (a) 1. Find the running speed for Marcia v M = 1.15(6.0 m/s) = 6.9 m/s 2. Find the time for Marcia t M = (100 m)/6.9 m/s) = 14.5 s 3. Find the distance covered by John in 14.5 s s J = (6.0 m/s)(14.5 s) = 87 m; distance = 13 m (b) Find the time required by John t J = (100 m)/(6 m/s) = 16.7 s; time difference = 2.2 s 14 Figure 2-22 shows the position of a particle versus time. Find the average velocities for the time intervals a, b, c, and d indicated in the figure. From the figure: s a = 0, so v av = 0; s b = 1 m, t b = 3 s, so v av = 0.33 m/s; s c = -6 m, t c = 3 s, so v av = -2 m/s; s d = 3 m, t d = 3 s, so v av = 1 m/s. 15 It has been found that galaxies are moving away from the earth at a speed that is proportional to their distance from the earth. This discovery is known as Hubble s law. The speed of a galaxy at distance r from the earth is given by v = Hr, where H is the Hubble constant, equal to 1.58 10-18 s -1. What is the speed of a galaxy (a) 5 10 22 m from earth and (b) 2 10 25 m from earth? (c) If each of these galaxies has traveled with constant speed, how long ago were they both located at the same place as the earth? (a) Use Hubble s law v a = (5 10 22 m)(1.58 10-18 s -1 ) = 7.9 10 4 m/s (b) Use Hubble s law v b = (2 10 25 m)(1.58 10-18 s -1 ) = 3.16 10 7 m/s (c) Use Equ. 2-3 for both galaxies t = r/v = r/rh = 1/H = 6.33 10 17 s = 2 10 10 y
16 Cupid fires an arrow that strikes St. Valentine, producing the usual sound of harp music and bird chirping as Valentine swoons into a fog of love. If Cupid hears these telltale sounds exactly one second after firing the arrow, and the average speed of the arrow was 40 m/s, what was the distance separating them? Take 340 m/s for the speed of sound. Let t 1 be the travel time of the arrow, and let t 2 be that of the sound. Both the sound and arrow travel a distance D. 1. Write expressions for D in terms of t 1 and t 2 D = (40 m/s)t 1 = (340 m/s)t 2 (1) 2. Write t 1 in terms of t 2 t 1 = (340/40)t 2 = 8.5t 2 (2) 3. The sum of t 1 and t 2 equals 1 s t 1 + t 2 = 1 s (3) 4. Solve (1) and (2) for t 1 and t 2 9.5t 2 = 1 s; t 2 = 0.105 s; t 1 = 0.895 s (4) 5. Use (1) to find D D = 35.8 m 17* If the instantaneous velocity does not change, will the average velocities for different intervals differ? No, they will not. For constant velocity, the instantaneous and average velocities are equal. 18 If v av = 0 for some time interval t, must the instantaneous velocity v be zero at some point in the interval? Support your answer by sketching a possible x-versus-t curve that has x = 0 for some interval t. Yes, it must. In the adjoining graph of x versus t, x = 0 in the interval between t = 0 and t = 4.0 s. Consequently, v av = x/ t = 0, although the instantaneous velocity is zero only at the point t = 2 s. 19 An object moves along the x axis as shown in Figure 2-23. At which point or points is the magnitude of its velocity at a minimum? (a) A and E (b) B, D, and E (c) C only (d) E only (e) None of these is correct. (b) At these points the slope of the position-versus-time curve is zero; therefore the velocity is zero. 20 For each of the four graphs of x versus t in Figure 2-24, answer the following questions. (a) Is the velocity at time t 2 greater than, less than, or equal to the velocity at time t 1? (b) Is the speed at time t 2 greater than, less than, or equal to the speed at time t 1? We shall use v to denote velocity and v to denote speed. (a) curve a: v(t 2 ) < v(t 1 ); curve b: v(t 2 ) = v(t 1 ); curve c: v(t 2 ) > v(t 1 ); curve d: v(t 2 ) < v(t 1 ). (b) curve a: v(t 2 ) < v(t 1 ); curve b: v(t 2 ) = v(t 1 ); curve c: v(t 2 ) < v(t 1 ); curve d: v(t 2 ) > v(t 1 ). 21* Using the graph of x versus t in Figure 2-25, (a) find the average velocity between the times t = 0 and t = 2 s. (b) Find the instantaneous velocity at t = 2 s by measuring the slope of the tangent line indicated. (a) Find x from graph; v av = x/ t x = 2 m, t = 2 s; v av = 1 m/s
(b) From graph, tangent passes through points x = 0, t = 1 s; x = 4 m, t = 3 s. Slope of tangent line is (4 m)/(2 s) = 2 m/s v(t = 2 s) = 2 m/s 22 Using the graph of x versus t in Figure 2-26, find (a) the average velocity for the time intervals t = t 2-0.75 s when t 2 is 1.75, 1.5, 1.25, and 1.0 s; (b) the instantaneous velocity at t = 0.75 s; (c) the approximate time when the instantaneous velocity is zero. (a) From graph find x(0.75), x(1.75), x(1.5), x(1.25), x(1.0) Find v av = x/ t (b) Draw tangent at t = 0.75 s and find slope (c) Find the value of t where slope is zero x(0.75) = 4.0 m, x(1.75) = 5.9 m, x(1.5) = 6 m, x(1.25) = 5.6 m, x(1.0) = 5 m 1.75 s, v av = 1.9 m/s; 1.5 s, v av = 2.67 m/s; 1.25 s, v av = 3.2 m/s; 1.0 s, v av = 4 m/s Slope = v(0.75) = 4.2 m/s t = 1.6 s 23 The position of a certain particle depends on the time according to x = (1 m/s 2 )t 2 - (5 m/s)t + 1 m. (a) Find the displacement and average velocity for the interval 3 s t 4 s. (b) Find the general formula for the displacement for the time interval from t to t + t. (c) Use the limiting process to obtain the instantaneous velocity for any time t. (a) 1. Find x(4) and x(3) x(4) = (16-20 + 1) m = -3 m; x(3) = (9-15 + 1) m = -5 m 2. Find x x = x(4) - x(3) = 2 m 3. Use Equ. 2-2 v av = x/ t = (2 m)/(1 s) = 2 m/s (b) 1. Find x(t + t) x(t + t) = [(t 2 + 2t t + t 2 ) - 5(t + t) + 1] m 2. Find x(t + t) - x(t) = x x = [(2t - 5) t + t 2 ] m (c) From (b) find x/ t as t 0 v = lim t 0 ( x/ t) = (2t - 5) m/s 24 The height of a certain projectile is related to time by y = -5(t - 5) 2 + 125, where y is in meters and t is in seconds. (a) Sketch y versus t for 0 t 10 s. (b) Find the average velocity for each of the 1-s time intervals between integral time values from 0 t 10 s. Sketch v av versus t. (c) Find the instantaneous velocity as a function of time. (a) The plot of y versus t is shown (b) Substitute t = 0,1,2,...,10 into the expression for y. The table below lists the values. Then evaluate y/ t = y/(1 s) to get v av, shown in the table. t y, m v av, m/s
0 0 0 1 45 45 2 80 35 3 105 25 4 120 15 5 125 5 6 120-5 7 105-15 8 80-25 9 45-35 10 0-45 (c) To find the instantaneous velocity, take the derivative of the expression for y(t); v(t) = dy/dt = (50-10t) m/s 25* The position of a body oscillating on a spring is given by x = A sin ωt, where A and ω are constants with values A = 5 cm and ω = 0.175 s -1. (a) Sketch x versus t for 0 t 36 s. (b) Measure the slope of your graph at t = 0 to find the velocity at this time. (c) Calculate the average velocity for a series of intervals beginning at t = 0 and ending at t = 6, 3, 2, 1, 0.5, and 0.25 s. (d) Compute dx/dt and find the velocity at time t = 0. (a) The plot of x versus t is shown (b) The slope of the dotted line (tangent at t = 0) is 0.875; v(0) = 0.875 cm/s (c) t x x/ t 6 4.34 0.723
3 2.51 0.835 2 1.71 0.857 1 0.174 0.871 0.5 0.437 0.874 0.25 0.219 0.875 (d) dx/dt = Aω cos ωt; at t= 0 cos ωt = 1; dx/dt at t = 0 is Aω = 0.875 cm/s 26 To avoid falling too fast during a landing, an airplane must maintain a minimum airspeed (the speed of the plane relative to the air). However, the slower the ground speed (speed relative to the ground) during a landing, the safer the landing. Is it safer for an airplane to land with the wind or against the wind? It is safer to land against the wind. 27 Two cars are traveling along a straight road. Car A maintains a constant speed of 80 km/h; car B maintains a constant speed of 110 km/h. At t = 0, car B is 45 km behind car A. How far will car A travel from t = 0 before it is overtaken by car B? 1. Find the velocity of car B relative to car A v rel = v B - v A = (110-80) km/h = 30 km/h 2. Find the time before overtaking t = s/v rel = (45 km)/(30 km/h) = 1.5 h 3. Find the distance traveled by car A in 1.5 h d = (1.5 h)(80 km/h) = 120 km 28 A car traveling at constant speed of 20 m/s passes an intersection at time t = 0, and 5 s later another car traveling 30 m/s passes the same intersection in the same direction. (a) Sketch the position functions x 1 (t) and x 2 (t) for the two cars. (b) Determine when the second car will overtake the first. (c) How far from the intersection will the two cars be when they pull even? (a) The plot of x 1 (t) and x 2 (t) are shown (b) From the plot, the second will overtake the first at t = 15 s (c) The two cars will be 300 m from the intersection
29* Margaret has just enough gas in her speedboat to get to the marina, an upstream journey that takes 4.0 hours. Finding it closed for the season, she spends the next 8.0 hours floating back downstream to her shack. The entire trip took 12.0 h; how long would it have taken if she had bought gas at the marina? Let D = distance to marina, v W = velocity of stream, v rel = velocity of boat under power relative to stream. 1. Express the times of travel with gas in terms of t 1 = D/(v rel - v W ) = 4 h D, v W, and v rel t 2 = D/(v rel + v W ) 2. Express the time required to drift distance D t 3 = D/v W = 8 h; v W = D/(8 h) 3. From t 1 = 4 h, find v rel 4(v rel - D/8) = D; v rel = 1.5D/(4 h) 4. Solve for t 2 t 2 = D/[(1.5D/4 h) + (D/8 h) = 2 h 5. Add t 1 and t 2 t tot = t 1 + t 2 = 6 h 30 Joe and Sally tend to argue when they travel. Just as they reached the moving sidewalk at the airport, their struggle for itinerary-making powers peaked. Though they stepped on the moving belt at the same time, Joe chose to stand and ride, while Sally opted to keep walking. Sally reached the end in 1 min, while Joe took 2 min. How long would it have taken Sally if she had walked twice as fast? Let v B = velocity of belt, v S velocity of Sally when walking normally, and D the length of the belt. 1. Write D in terms of v B ; Joe s speed on the belt D = (2 min)(v B ); v B = D/(2 min) 2. Write D in terms of v B + v S ; Sally s speed on the D = (1 min)(v B + v S ) = (1 min)[(d/2 min) + v S ] belt 3. Solve for v S v S = D/(2 min) 4. Write t 2 = time for fast walk t 2 = D/(v B + 2v S ) = D/[(D/2 min) + (D/1 min) = 40 s 31 Walk across the room in such a way that, after getting started, your velocity is negative but your acceleration is positive. (a) Describe how you did it. (b) Sketch a graph of v versus t for your motion. (a) Start walking with velocity in the negative direction. Gradually slow the speed of walking, until the other end of the room is reached. (b) A sketch of v versus t is shown 32 Give an example of a motion for which both the acceleration and the velocity are negative. Let up be the positive direction. A falling object then has a negative velocity and a negative acceleration. 33* Is it possible for a body to have zero velocity and nonzero acceleration? Yes, it is. An object tossed up has a constant downward acceleration; at its maximum height, its instantaneous velocity is zero. 34 True or false: (a) If the acceleration is zero, the body cannot be moving. (b) If the acceleration is zero, the x-versus-t curve must be straight line. (a) False (b) True
35 State whether the acceleration is positive, negative, or zero for each of the functions x(t) in Figure 2-27. (a) a = 0, constant velocity. (b) a > 0, v changes from negative to positive. (c) a < 0. (d) a = 0. 36 Answer the following question for each of the graphs in Figure 2-28: (a) At what times are the accelerations of the objects positive, negative, and zero? (b) At what times are the accelerations constant? (c) At what times are the instantaneous velocities zero? Curve 1: v versus t. Hence, a < 0 for t < 3 s and t > 7 s; a = 0 for t = 3 s and 6 s t 7 s; a > 0 for 3 s t 6 s. (b) a is constant for t < 2.7 s, 3.2 s t 5.8 s, 6 s t 7 s, 7.1 s t. (c) v = 0 at t = 8.6 s. Curve 2: x versus t. Here, a < 0 for 0 t 3 s and t > 7 s; a = 0 for 3 s t 5 s; a > 0 for 5 s t 7 s. (b) It is difficult to tell where a is constant from the curve; if the curved segments are parabolas, then a is constant everywhere. (c) v = 0 at t = 2 s, 5.8 s, and 7.8 s. 37* A BMW M3 sports car can accelerate in third gear from 48.3 km/h (30 mi/h) to 80.5 km/h (50 mi/h) in 3.7 s. (a) What is the average acceleration of this car in m/s 2? (b) If the car continued at this acceleration for another second, how fast would it be moving? (a) 1. Use Equ. 2-8 a av = [(80.5-48.3) km/h]/(3.7 s) = 8.7 km/h.s 2. Convert to m/s 2 (8.7 10 3 m/h s)(1 h/3600 s) = 2.42 m/s 2 (b) In 1 s, its speed increases by 8.7 km/h v = (80.5 + 8.7) km/h = 89.2 km/h 38 At t = 5 s, an object at x = 3 m is traveling at 5 m/s. At t = 8 s, it is at x = 9 m and its velocity is -1 m/s. Find the average acceleration for this interval. Use Equ. 2-8, a av = v/ t a av = [(-1 m/s) - (5 m/s)]/[(8 s) - (5 s)] = -2 m/s 2 39 A particle moves with velocity v = 8t - 7, where v is in meters per second and t is in seconds. (a) Find the average acceleration for the one-second intervals beginning at t = 3 s and t = 4 s. (b) Sketch v versus t. What is the instantaneous acceleration at any time? (a) 1. Determine v at t = 3 s, t = 4 s, t = 5 s 2. Find a av for the two 1-s intervals (b) v versus t is shown in the adjacent plot Use Equ. 2-10: a = dv/dt = 8 m/s 2. v(3) = 17 m/s; v(4) = 25 m/s; v(5) = 33 m/s a av (3-4) = 8/1 m/s 2 = 8 m/s 2 ; a av (4-5) = 8 m/s 2 40 The position of an object is related to time by x = At 2 - Bt + C, where A = 8 m/s 2, B = 6 m/s, and C = 4 m. Find the instantaneous velocity and acceleration as functions of time. 1. Use Equs. 2-5 and 2-10 v = dx/dt = 2At - B; a = dv/dt = 2A
2. Substitute numerical values for A and B v = (16t - 6) m/s; a = 16 m/s 2 41* Identical twin brothers standing on a bridge each throw a rock straight down into the water below. They throw rocks at exactly the same time, but one hits the water before the other. How can this occur if the rocks have the same acceleration? The initial downward velocities of the two rocks are not the same. 42 A ball is thrown straight up. What is the velocity of the ball at the top of its flight? What is its acceleration at that point? At the top of its flight, its velocity is instantaneously zero; its acceleration is 9.81 m/s 2 downward. 43 An object thrown straight up falls back to the ground T seconds later. Its maximum height is H meters. Its average velocity during those T seconds is (a) H/T, (b) 0, (c) H/2T, (d) 2H/T. (b) The displacement is zero. 44 For an object thrown straight up, which of the following is true while it is in the air? (a) The acceleration is always opposite to the velocity. (b) The acceleration is always directed downward. (c) The acceleration is always in the direction of motion. (d) The acceleration is zero at the top of the trajectory. (b) The acceleration is -g, always directed downward. 45* An object projected up with initial velocity v attains a height H. Another object projected up with initial velocity 2v will attain a height of (a) 4H, (b) 3H, (c) 2H, (d) H. (a) 4H; from Equ. 2-15, with a = -g and v = 0 at top of trajectory, H = v0 2 / 2g, v 0 is initial velocity. So H v 2 0. 46 A ball is thrown upward. While it is in the air, its acceleration is (a) decreasing, (b) constant, (c) zero, (d) increasing. (b) constant; downward with value g. 47 At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the roof. During their descent to the ground the distance between the two objects (a) is proportional to t. (b) is proportional to t 2. (c) decreases. (d) remains 10 m throughout. (d) Both move with the same acceleration and velocity at all times; the distance between them remains that at t = 0. 48 A Porsche accelerates uniformly from 80.5 km/h (50 mi/h) at t = 0 to 113 km/h (70 mi/h) at t = 9 s. Which graph in Figure 2-29 best describes the motion of the car? (c) v(0) > 0 and v(t) increases at a uniform rate. 49* An object is dropped from rest. If the time during which it falls is doubled, the distance it falls will (a) double, (b) decrease by one-half, (c) increase by a factor of four, (d) decrease by a factor of four, (e) remain the same. (c) increase by a factor of 4; see Equ. (2-14): x t 2 if v 0 = 0. 50 A ball is thrown upward with an initial velocity v 0. Its velocity halfway to its highest point is (a) 0.5v 0, (b) 0.25v 0, (c) v 0, (d) 0.707v 0, (e) cannot be determined from the information given. (d) 0.707v 0 ; H = v 2 0 /2g, v 2 = v 2 0-2g(H/2) = 1/2v 2 0 ; v = 0.707v 0. 51 A car starting at x = 50 m accelerates from rest at a constant rate of 8 m/s 2. (a) How fast is it going after 10 s? (b) How far has it gone after 10 s? (c) What is its average velocity for the interval 0 t 10 s?
(a) Use Equ. 2-12 (b) Use Equ. 2-14 (c) Use Equ. 2-2 v = 0 + (8 m/s 2 )(10 s) = 80 m/s x = 1/2(8 m/s 2 )(10 s) 2 = 400 m v av = (400 m)/(10 s) = 40 m/s 52 An object with an initial velocity of 5 m/s has a constant acceleration of 2 m/s 2. When its speed is 15 m/s, how far has it traveled? Use Equ. 2-15 x = [(15 2-5 2 ) m 2 /s 2 ]/[2(2 m/s 2 )] = 50 m 53* An object with constant acceleration has velocity v = 10 m/s when it is at x = 6 m and v = 15 m/s when it is at x = 10 m. What is its acceleration? Use Equ. 2-15 a = [(15 2-10 2 ) m 2 /s 2 ]/[2(4 m)] = 15.6 m/s 2 54 An object has constant acceleration a = 4 m/s 2. At t = 0, its velocity is 1 m/s and it is at x = 7 m. How fast is it moving when it is at x = 8 m? What is t at that point? 1. Use Equ. 2-15 v 2 = 1 m 2 /s 2 + 2(4 m/s 2 )(1 m) = 9 m 2 /s 2 ; v = 3 m/s 2. Use Equ. 2-12 t = [(3-1) m/s]/(4 m/s 2 ) = 0.5 s 55 If a rifle fires a bullet straight up with a muzzle speed of 300 m/s, how high will the bullet rise? (Ignore air resistance.) At top, v = 0; use Equ. 2-15 H = v 2 0 /2g = 4.59 10 3 m 56 A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 98 km/h to zero is 50 m. Find the acceleration, assuming it to be constant, and express your answer as a fraction of the free-fall acceleration due to gravity. How long does the car take to stop? 1. Use Equ. 2-15 a = -(98 km/h) 2 /(100 m) = -9.6 10 7 m/h 2 = -7.41 m/s 2 2. Divide a by g = -9.81 m/s 2 a = 0.755g 3. Use Equ. 2-12 t = (98 km/h)/(9.6 10 7 m/h 2 ) = 1.02 10-3 h = 3.67 s 57* A ball is thrown upward with an initial velocity of 20 m/s. (a) How long is the ball in the air? (b) What is the greatest height reached by the ball? (c) When is the ball 15 m above the ground? (a) 1. Take upward as positive; use Equ. 2-14 2. Solve for t (b) See Problem 55 (c) 1. Use Equ. 2-14 -b ± 2. Use quadratic formula t = 2 b 2a x = 0 = (20 m/s)t - 1/2(9.81 m/s 2 )t 2 t = 0; t = 4.08 s; t = 4.08 s is the proper result H = (400 m 2 /s 2 )/[2(9.81 m/s 2 )] = 20.4 m 15 m = (20 m/s)t - 1/2(9.81 m/s 2 )t 2-4ac t = 0.991 s, t = 3.09 s; both are acceptable solutions 58 A particle moves with constant acceleration of 3 m/s 2. At t = 4 s, it is at x = 100 m; at t = 6 s, it has a velocity v = 15 m/s. Find its position at t = 6 s. 1. Find v(4 s) = v 0 using Equ. (2-12) v 0 = (15-2 3) m/s = 9 m/s 2. Use Equ. 2-14; elapsed time = 2 s x = [100 m + (9 m/s)(2 s) + 1/2(3 m/s 2 )(2 s) 2 ] = 124 m
59 A bullet traveling at 350 m/s strikes a telephone pole and penetrates a distance of 12 cm before stopping. (a) Estimate the average acceleration by assuming it to be constant. (b) How long did it take for the bullet to stop? (a) Use Equ. 2-15 a = -(350 m/s) 2 /[2(0.12 m)] = -5.1 10 5 m/s 2 (b) Use Equ. 2-12 t = (350 m/s)/(5.1 10 5 m/s 2 ) = 0.686 ms 60 A plane landing on an aircraft carrier has just 70 m to stop. If its initial speed is 60 m/s, (a) what is the acceleration of the plane during landing, assuming it to be constant? (b) How long does it take for the plane to stop? (a) Use Equ. 2-15 a = -(60 m/s) 2 /[2(70 m)] = -25.7 m/s 2 (b) Use Equ. 2-12 t = (60 m/s)/(25.7 m/s 2 ) = 2.33 s 61* An automobile accelerates from rest at 2 m/s 2 for 20 s. The speed is then held constant for 20 s, after which there is an acceleration of -3 m/s 2 until the automobile stops. What is the total distance traveled? 1. Determine the distance traveled during first 20 s and the speed at the end of first 20 s 2. Find x 2 = distance covered 20 s t 40 s 3. Find x 3 = distance during deceleration 4. Find total distance v(20) = at = (2 m/s 2 )(20 s) = 40 m/s x 1 = v av t = (20 m/s)(20 s) = 400 m x 2 = (40 m/s)(20 s) = 800 m x 3 = (40 m/s) 2 /[2(3 m/s 2 )] = 267 m x = x 1 + x 2 + x 3 = 1467 m 62 In the Blackhawk landslide in California, a mass of rock and mud fell 460 m down a mountain and then traveled 8 km across a level plain on a cushion of compressed air. Assume that the mud dropped with the free-fall acceleration due to gravity and then slid horizontally with constant deceleration. (a) How long did the mud take to drop the 460 m? (b) How fast was it traveling when it reached the bottom? (c) How long did the mud take to slide the 8 km horizontally? (a) Use Equ. 2-14 (b) Use Equ. 2-12 (c) Use Equ. 2-13 t 2 = 2(460 m)/(9.81 m/s 2 ) = 93.8 s 2 ; t = 9.68 s v = (9.81 m/s 2 )(9.68 s) = 95.0 m/s t = 2(8000 m)/(95.0 m/s) = 168 s 63 A load of bricks is being lifted by a crane at a steady velocity of 5 m/s when one brick falls off 6 m above the ground. (a) Sketch x(t) to show the motion of the free brick. (b) What is the greatest height the brick reaches above the ground? (c) How long does it take to reach the ground? (d) What is its speed just before it hits the ground? (a) Use Equ. 2-14; x 0 = 6 m, v 0 = 5 m/s, a = -9.81 m/s 2 (b) Use Equ. 2-15; find x and add to x 0 ; x = (25/2 9.81) m = 1.27 m; x = 7.27 m (also see graph) (c) Use Equ. 2-14; then solve quadratic equ. for t
0 = 6 + 5t - 1/2 9.81t 2 ; t = 1.73 s, t = -0.708 s Second (negative solution) is non-physical; t = 1.73 s (d) Use v 2 = 2gH, with H = 7.27 m v 2 = 2 9.81 7.27 m 2 /s 2 ; v = 11.9 m/s 64 An egg with a mass of 50 g rolls off a table at a height of 1.2 m and splatters on the floor. Estimate the average acceleration of the egg while it is in contact with the floor. We shall take 5 cm as the diameter of an egg, and consider the time of contact while crashing, during which the egg travels 5 cm. Just before contact with the floor, v = 2gH» 5 m/s, so the time of contact» (0.05/2.5) s = 20 ms. In that time the egg decelerates to v = 0. So the magnitude of a av = (5/0.02) m/s 2 = 250 m/s 2. 65* To win publicity for her new CD release, Sharika, the punk queen, jumps out of an airplane without a parachute. She expects a stack of loose hay to break her fall. If she reaches a speed of 120 km/h prior to impact, and if a 35 g deceleration is the greatest deceleration she can withstand, how high must the stack of hay be in order for her to survive? Assume uniform acceleration while she is in contact with the hay. 1. Use Equ. 2-15 5 2 (1.20 x10 m/h ) H = = 1.62 m 2 2 (3600 s/1 h) [2x35x(9.81 m/ s )] 66 A bolt comes loose from underneath an elevator that is moving upward at a speed of 6 m/s. The bolt reaches the bottom of the elevator shaft in 3 s. (a) How high up was the elevator when the bolt came loose? (b) What is the speed of the bolt when it hits the bottom of the shaft? We use up as the positive direction; x = 0 at bottom of shaft. (a) Use Equ. 2-14; v 0 = 6 m/s, t = 3 s, x(3 s) = 0 x 0 =[ 0 - (6 3) - 1/2(-9.81)(9)] m = 26.1 m (b) Use Equ. 2-12 v = [6-9.81 3] m/s = -23.4 m/s; v = 23.4 m 67 An object is dropped from a height of 120 m. Find the distance it falls during its final second in the air. 1. Find final (impact) speed, v f 2. Find v f-1, speed 1 s prior to impact 3. Find the average speed during this 1 s 4. Find the distance traversed, s = v av t v f = 2 (9.81) (120) m/s = 48.5 v f-1 = (48.5-9.81) m/s = 38.7 m/s v av = 1/2(48.5 + 38.7) m/s = 43.6 m/s s = (43.6 1) m = 43.6 m 68 An object is dropped from a height H. During the final second of its fall, it traverses a distance of 38 m. What was H? Let v f be the final speed before impact, v f-1 the speed 1 s before impact. 1. Find the average speed in last second 2. Express v f-1 in terms of v f and solve for v f 3. Use v f to determine H m/s v av (1 s) = 38 m = 1/2(v f + v f-1 )(1 s); v f + v f-1 = 76 m/s v f - v f-1 = 9.81 m/s; v f = 42.9 m/s H = v f 2 /2g = 93.8 m 69* A stone is thrown vertically from a cliff 200 m tall. During the last half second of its flight the stone travels a distance of 45 m. Find the initial velocity of the stone.
We take down as the positive direction. Let v 1 be the velocity 1/2 s before impact, v f velocity at impact. 1. Find v av during last 1/2 s 2. Write v f in terms of v 1 and g 3. Solve for v f v av = 1/2(v 1 + v f ) = (45/0.5) m/s; v 1 + v f = 180 m/s v f = v 1 + gt; v f = v 1 + (9.81 0.5) m/s v f = 92.5 m/s 4. Use Equ. 2-15 to find v 0 = v0 ( 92.5) 2-2x9.81x200 m/s = ± 68 m/s ; the stone may be thrown either up or down 70 An object in free-fall from a height H traverses 0.4H during the first second of its descent. Determine the average speed of the object during free-fall. 1. Find x for first second and H 2. Find v f 3. Determine v av, v 0 = 0 x = 1/2gt 2 = 4.9 m = 0.4H; H = 12.3 m v f = [2(9.81)(12.3)] 1/2 m/s = 15.5 m/s v av = 1/2v f = 7.77 m/s 71 A bus accelerates at 1.5 m/s 2 from rest for 12 s. It then travels at constant speed for 25 s, after which it slows to a stop with an acceleration of -1.5 m/s 2. (a) How far did the bus travel? (b) What was its average velocity? (a) 1. Find v(12) and v av (0-12) 2. Find x(12) = v av (0-12) t 3. Find x(12-37) = v(12) t 4. deceleration = acceleration; x dec = x acc 5. Add displacements to find x tot v(12) = (1.5 12) m/s = 18 m/s; v av (0-12) = 9 m/s x(12) = (9 12) m = 108 m x(12-37) = (18 25) m = 450 m x dec = 108 m and t dec = 12 s Total distance = 666 m (b) v av = x tot /t tot v av = (666 m)/(49 s) = 13.6 m/s 72 A basketball is dropped from a height of 3 m and rebounds from the floor to a height of 2 m. (a) What is the velocity of the ball just as it reaches the floor? (b) What is its velocity just as it leaves the floor? (c) Estimate the magnitude and direction of its average acceleration during this interval. (a) v = 2 gh v = -(2 9.81 3) 1/2 m/s = -7.67 m/s v = +(2 9.81 2) 1/2 m/s = 6.26 m/s (b) Use the above expression a av = (6.26 + 7.67)/0.05 m/s 2 = 279 m/s 2 upward (c) A reasonable estimate for t is 0.05 s 73* A rocket is fired vertically with an upward acceleration of 20 m/s 2. After 25 s, the engine shuts off and the rocket continues as a free particle until it reaches the ground. Calculate (a) the highest point the rocket reaches, (b) the total time the rocket is in the air, (c) the speed of the rocket just before it hits the ground. Take up as the positive direction. (a) 1. Find x 1 and v 1 at t = 25 s; use Equ. 2-14 2. Find x 2 = distance above x 1 when v = 0 3. Total height = x 1 + x 2 (b) 1. Find time, t 2, for part (a) 2. Find time, t 3, to drop 19.0 km 3. Total time, T = 25 s + t 2 + t 3 (c) Use Equ. 2-12 x 1 = 1/2(20)(25) 2 m = 6250 m; v 1 = 20 25 m/s = 500 m/s x 2 = (500) 2 /(2 9.81) m = 1.274 10 4 m H = 1.90 10 4 m = 19.0 km t 2 = x 2 /v av = (1.274 10 4 /250) s = 51 s t 3 = [2(1.90 10 4 )/9.81] 1/2 s = 62.2 s T = (25 + 51 + 62.2) s = 138 s = 2 min 18 s v f = (9.81)(62.2) m/s = 610 m/s
74 A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally holding a stopwatch, notices that it takes 0.2 s for the pot to fall past his window, which is 4 m high. How far above the top of the window is the ledge from which the pot fell? Take down as positive, x = 0 at ledge. Let t = time when pot is at top of window, t + t when pot is at bottom of window. 1. Express x = 4 m in terms of t, t, and g 4 m = 1/2g[(t+ t) 2 - t 2 ] = 1/2g( t) 2 + gt t 2. Solve for t, with t = 0.2 s, g = 9.81 m/s 2 t = 1.94 s 3. Find distance pot falls from rest in 1.94 s H = 1/2(9.81)(1.94) 2 = 18.4 m 75 Sharika arrives home late from a gig, only to find herself locked out. Her roommate and bass player Chico is practicing so loudly that he can t hear Sharika s pounding on the door downstairs. One of the band s props is a small trampoline, which Sharika places under Chico s window. She bounces progressively higher trying to get Chico s attention. Propelling herself furiously upward, she miscalculates on the last bounce and flies past the window and out of sight. Chico sees her face for 0.2 s as she moves a distance of 2.4 m from the bottom to the top of the window. (a) How long until she reappears? (b) What is her greatest height above the top of the window? (Treat Sharika as a point-particle punk.) Take up as positive; x = 0 at bottom of window, v 0 = upward velocity at top of window. (a) 1. Use Equ. 2-14 and solve for v 0 2.4 m = v 0 (0.2 s) + 1/2(9.81)(0.2) 2 m; v 0 = 11.0 m/s 2. Find the time to reach v = 0 t = (11.0/9.81) s = 1.12 s 3. Time when her face reappears is 2t Time till her face reappears is 2.24 s (b) Find H using Equ. 2-15 H = v 2 0 /2g = 6.17 m 76 In a classroom demonstration, a glider moves along an inclined air track with constant acceleration a. It is projected from the start of the track (x = 0) with an initial velocity v 0. At time t = 8 s, it is at x = 100 cm and is moving along the track at velocity v = -15 cm/s. Find the initial speed v 0 and the acceleration a. 1. Use Equ. (2-13) and solve for v 0 v 0 = 2x/t - v = [(200/8) - (-15)] m/s = 40 cm/s 2. Use the definition of a = v/ t a = (-55/8) cm/s 2 = -6.88 cm/s 2 77* A rock dropped from a cliff falls one-third of its total distance to the ground in the last second of its fall. How high is the cliff? 1. Write the final speed in terms of H and g v 2 f = 2gH 2. Write the average speed in the last second v av =1/2[v f + (v f - g)] =(v f - g/2) 3. Set the distance in last second = H/3; solve for v f (v f - g/2) = H/3; v f = H/3 + g/2 4. Set v 2 f = 2gH and solve for H 2gH = H 2 /9 + gh/3 + g 2 /4; H = 14.85g = 145.7 m 78 A typical automobile has a maximum deceleration of about 7 m/s 2 ; the typical reaction time to engage the brakes is 0.50 s. A school board sets the speed limit in a school zone to meet the condition that all cars should be able to stop in a distance of 4 m. (a) What maximum speed should be allowed for a typical automobile? (b) What fraction of the 4 m is due to the reaction time? (a) 1. Write the total distance as the sum of distance x 1 = 0.50v 0 ; x 2 = -v 0 2 /2a (a = -7 m/s 2 )
traveled in 0.50 s plus that during deceleration 2. Solve quadratic equ. keeping positive result (b) Find reaction time distance 4 m = x 1 + x 2 = (0.5v 0 + v 0 2 /14) m v 0 = 4.76 m/s = 4.76 m/s = 10.6 mi/h x 1 = 2.38m = (2.38/4)? x tot = 59.5% of )? x tot 79 Two trains face each other on adjacent tracks. They are initially at rest 40 m apart. The train on the left accelerates rightward at 1.4 m/s 2. The train on the right accelerates leftward at 2.2 m/s 2. How far does the train on the left travel before the two trains pass? Take x = 0 as position of train on left at t = 0. 1. Write positions of each train as a function of time x L = 0.7t 2 m; x R = 40-1.1t 2 2. Set x L = x R and solve for t 0.7t 2 = 40-1.1t 2 ; t = 4.71 s 3. Find distance traveled, i.e., x L x L = 15.6 m 80 Two stones are dropped from the edge of a 60-m cliff, the second stone 1.6 s after the first. How far below the cliff is the second stone when the separation between the two stones is 36 m? 1. Write expressions for x 1 and x 2 x 1 = 1/2gt 2 ; x 2 = 1/2g(t - 1.6 s) 2 2. Set x 1 - x 2 = 36 m 36 m = 1/2g[(3.2 s)t - 2.56 s 2 ] 3. Solve for t, then find x 2 t = 3.09 s; x 2 = 10.9 m 81* A motorcycle policeman hidden at an intersection observes a car that ignores a stop sign, crosses the intersection, and continues on at constant speed. The policeman takes off in pursuit 2.0 s after the car has passed the stop sign, accelerates at 6.2 m/s 2 until his speed is 110 km/h, and then continues at this speed until he catches the car. At that instant, the car is 1.4 km from the intersection. How fast was the car traveling? 1. Find the time of travel of policeman: t 1 is the time t 1 = (1.10 10 3 m/h)/[(3600 s/1 h)(6.2 m/s 2 )] = 4.93 s of acceleration, t 2 the time of travel at 110 km/h; d 1 = 1/2(30.5 m/s)(4.93 s) = 75.3 m; d 2 = (1400-75.3) m and d 1 and d 2 the corresponding distances = 1325 m; t 2 = d 2 /(30.6 m/s) = 43.3 s 2. The time of travel of car is 2.0 s + t 1 + t 2 t C = (2.0 + 43.3 + 4.93) s = 50.2 s 3. Find the speed of the car v C = (1400/50.2) m/s = 27.9 m/s = 100.4 km/h 82 At t = 0, a stone is dropped from a cliff above a lake; 1.6 seconds later another stone is thrown downward from the same point with an initial speed of 32 m/s. Both stones hit the water at the same instant. Find the height of the cliff. 1. Write the distances dropped and equate 1/2gt 2 = (32 m/s )(t - 1.6 s) + 1/2g(t - 1.6 s) 2 2. Solve for t t = 2.37 s 3. Find H; Use Equ. 2-14 H = 1/2gt 2 = 27.6 m 83 A passenger train is traveling at 29 m/s when the engineer sees a freight train 360 m ahead traveling on the same track in the same direction. The freight train is moving at a speed of 6 m/s. If the reaction time of the engineer is 0.4 s, what must be the deceleration of the passenger train if a collision is to be avoided? If your answer is the maximum deceleration of the passenger train but the engineer s reaction time is 0.8 s, what is the relative speed of the two trains at the instant of
collision and how far will the passenger train have traveled in the time between the sighting of the freight train and the collision? Let x = 0 be the location of the passenger train engine at moment of sighting of freight train s end; let t = 0 be the instant the passenger train decelerates. 1. Write expressions for x P and x F, positions of the x P = [(29 m/s)(t + 0.4 s) - 1/2at 2 ]; x P = 11.6 + 29t -4.91t 2 passenger train engine and the freight train s end x F = (360 m) + (6 m/s)(t + 0.4 s); x F = 362.4 + 6t 2. Set x F = x P and obtain an equation for t 1/2at 2 - (23 m/s)t + 351 m = 0 3. If equation has a real solution collision occurs If (23 2-702a) > 0, collision 4. Solve for a so there is no real solution for t a 0.754 m/s 2 5. Repeat, with a = 0.754 m/s 2 and 0.8 s reaction quadratic equation is 0.377t 2-23t + 342 = 0 time 6. Solve for t, keeping only smaller value t = 25.5 s; (t = 35.5 s: trains have already collided) 7. Find x P at that instant x P = (29 m/s)(26.3 s) - ( 0.377 m/s 2 )(25.5 s) 2 = 518 m 8. Find the speeds of the two trains v P = (29 m/s) - (0.754 m/s 2 )(25.5 s) = 9.77 m/s; v F = 6 m/s; v rel = 3.77 m/s The adjacent figure shows the location of the trains. The solid straight line is for the freight train; the solid and dashed curved lines are for the passenger train, with reaction times of 0.4 s and 0.8 s, respectively. 84 After being forced out of farming, Lou has given up on trying to find work locally and is about to ride the rails to look for a job. Running at his maximum speed of 8 m/s, he is a distance d from the train when it begins to accelerate from rest at 1.0 m/s 2. (a) If d = 30 m and Lou keeps running, will he be able to jump into the train? (b) Sketch the position function x(t) for the train, with x = 0 at t = 0. On the same graph, sketch x(t) for Lou for various distances d, including d = 30 m and the critical separation distance d c, the distance at which he just catches the train. (c) For the situation d = d c, what is the speed of the train when Lou catches it? What is the train s average speed for the time interval between t = 0 and the moment Lou catches the train? What is the exact value of d c? (a), (b) and (c) For the train, x T (t) = 1/2(1.0 m/s 2 )t 2 ; for Lou, x L (t) = (8 m/s)t - d (a) Yes, he can jump on the train. (see graph) (c) When Lou just manages to catch the train, his speed and that of the train are equal. The speed of the train is v T (t) = 8
m/s = at = t m/s. Hence the critical time is t C = 8 s. At that time, Lou has run 64 m and the train has traveled 32 m. Consequently, d C = (64-32) m = 32 m. v av = (32/8) m/s = 4 m/s. The three straight lines on the graph correspond to d = 30 m, d = d C = 32 m, and d = 40 m. 85* A train pulls away from a station with a constant acceleration of 0.4 m/s 2. A passenger arrives at the track 6.0 s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and catch the train? Sketch curves for the motion of passenger and the train as functions of time. 1. As in Problem 84, the critical conditions are v T = 0.4t m/s; x T = 0.2t 2 m; x P = v P (t - 6 s) v T = v P and x T = x P 0.2t 2 = 0.4t(t - 6) 2. Solve for t 0.2t 2 = 2.4t; t = 12 s 3. Find v T at t = 12 s, which is also v PC v PC = 4.8 m/s The positions of the train and passenger as functions of time are shown in the adjoining figure. 86 Lou applies for a job as a perfume salesman. He tries to convince the boss to try his daring, aggressive promotional gimmick: dousing prospective customers as they wait at bus stops. A hard ball is to be thrown straight upward with an initial speed of 24 m/s. A thin-skinned ball filled with perfume is then thrown straight upward along the same path with a speed of 14 m/s. The balls are to collide when the perfume ball is at the high point of its trajectory, so that it breaks open
and everyone gets a free sample. If t = 0 when the first ball is thrown, find the time when the perfume ball should be thrown. 1. Find t for the perfume ball to reach the top v = 0 = (14-9.81 t) m/s; t = 1.427 s 2. Find height at t = 1.427 s H = (14 m/s) 2 /2g = 9.99 m 3. Find the time for the hard ball to be at H on its 9.99 = 24t - 1/2(9.81)t 2 ; solve for t downward motion t = 4.43 s (t = 0.46 s corresponds to upward motion) 4. Time delay is t - t Throw the perfume ball at t = (4.43-1.43) s = 3 s 87 Ball A is dropped from the top of a building at the same instant that ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what fraction of the height of the building does the collision occur? Take x = 0 at ground, upward positive. 1. Describe the conditions at collision x A = x B ; v A = -2v B (Note: A is moving down) 2. Write expressions for x A and x B 3. Write expressions for v A and v B 4. Set v A = -2v B and solve for t 5. Use this result in part 2 with x A = x B 6. Write x A (see part 2) in terms of v 0 x A = H - 1/2gt 2 ; x B = v 0 t - 1/2gt 2 v A = -gt; v B = v 0 - gt t = 2v 0 /3g H = 2v 2 0 /3g x A = 4v 2 0 /9g = 2H/3 88 Solve Problem 87 if the collision occurs when the balls are moving in the same direction and the speed of A is 4 times that of B. Proceed as in Problem 87 except that now the condition on velocities is v A = 4v B. One obtains for time t = 4v 0 /3g, and for velocities v A = -(4/3)v 0 and v B = -(1/3)v 0. Now H = 4v 2 0 /3g, giving x A = H/3. 89* The Sprint missile, designed to destroy incoming ballistic missiles, can accelerate at 100g. If an ICBM is detected at an altitude of 100 km moving straight down at a constant speed of 3 10 4 km/h and the Sprint missile is launched to intercept it, at what time and altitude will the interception take place? (Note: You can neglect the acceleration due to gravity in this problem. Why?) 1. Neglect g; see below. Find x ICBM, x S x S = 1/2at 2 ; x ICBM = H - vt 2. Express v in m/s v = (3 10 7 m/h)(1 h/3600 s) = 8.33 10 3 m/s 3. Set x ICBM = x S with H = 10 5 m and find t 1/2 981t 2 + 8.33 10 3 t - 10 5 = 0; t = 8.12 s 4. Find x ICBM x ICBM = [981 8.12 2 /2] m = 3.24 10 4 m = 32.4 km Note: In 8 s, the change in velocity v of the ICBM due to g is less than 80 m/s, i.e., less than 1% of v; also, g = 1% of a S. So the result is good to about 1%. Also, if a S = 100g is taken to be the possible horizontal acceleration, then both objects suffer the same downward acceleration due to gravity, and this contribution cancels. 90 When a car traveling at speed v 1 rounds a corner, the driver sees another car traveling at a slower speed v 2 a distance d ahead. (a) If the maximum acceleration the driver s brakes can provide is a, show that the distance d must be greater than (v 1 - v 2 ) 2 /(2a) if a collision is to be avoided. (b) Evaluate this distance for v 1 = 90 km/h, v 2 = 45 km/h, and a = 6 m/s 2. (c) Estimate or measure your reaction time and calculate the effect it would have
on the distance found in part (b). Note: This problem is similar to Problem 83. Again, the critical conditions are x 1 = x 2 and v 1 = v 2. (a) 1. Write expressions for x 1, x 2, v 1, and v 2 2. Set v 1 (t) = v 2 3. Set x 1 = x 2 and solve for d (b) Convert km/h to m/s and substitute to find d (c) Assume reaction time of 0.8 s. In that time, the distance between cars diminishes by 10 m. x 1 = v 1 t - 1/2at 2 ; x 2 = d + v 2 t; v 1 (t) = v 1 - at; v 1 = v 2 t = (v 1 - v 2 )/a d = (v 1 - v 2 ) 2 /(2a) d = [(25 m/s) - (12.5 m/s)] 2 /[2(6 m/s 2 )] = 13.0 m d = (13.0 + 10) m = 23 m 91 The velocity of a particle is given by v = 6t + 3, where t is in seconds and v is in meters per second. (a) Sketch v(t) versus t, and find the area under the curve for the interval t = 0 to t = 5 s. (b) Find the position function x(t). Use it to calculate the displacement during the interval t = 0 to t = 5 s. (a) The graph is shown. The area under the straight line is 90 m. (b) x(t) = t 0 (6t + 3)dt = 3t2 + 3t; x = x(5) - x(0) = 90 m. 92 Figure 2-30 shows the velocity of a particle versus time. (a) What is the magnitude in meters of the area of the rectangle indicated? (b) Find the approximate displacement of the particle for the one-second intervals beginning at t = 1 s and t = 2 s. (c) What is the approximate average velocity for the interval 1 s t 3 s? (a) Find the area of the rectangle A = (1 m/s)(1 s) = 1 m (b) 1. Find approx. area under curve x(1-2)» 1 m 2. Find approx. area under curve x(2-3)» 3 m (c) v av = x/ t v av» (4/2) m/s = 2 m/s 93* The velocity of a particle is given by v = 7t 2-5, where t is in seconds and v is in meters per second. Find the general position function x(t). 2 3 x(t) = ( 7t - 5 )dt = ( 7/3)t - 5t + C. 94 The equation of the curve shown in Figure 2-30 is v = 0.5t 2 m/s. Find the displacement of the particle for the
interval 1 s t 3 s by integration, and compare this answer with your answer for Problem 92. Is the average velocity equal to the mean of the initial and final velocities for this case? 1. x = 3 1 0.5t 2 dt = (1/6) t 3 3 = 4.33 m. 1 2. a = dv/dt = 1.0t m/s 2 ; a is not constant, therefore v av v mean. 95 Figure 2-31 shows the acceleration of a particle versus time. (a) What is the magnitude of the area of the rectangle indicated? (b) The particle starts from rest at t = 0. Find the velocity at t = 1 s, 2 s, and 3 s by counting the rectangles under the curve. (c) Sketch the curve v(t) versus t from your results for part (b), and estimate how far the particle travels in the interval t = 0 to t = 3 s. (a) Find the area of the rectangle A = (0.5 m/s 2 )(0.5 s) = 0.25 m/s (b) v = 0 at t = 0; count squares and multiply by v(1) = 0.9 m/s, v(2) = 3 m/s, v(3) = 6 m/s 0.25 (c) The curve of v(t) versus t is shown. By counting squares under the curve, we find the distance traveled is approximately 6.5 m 96 Figure 2-32 is a graph of v versus t for a particle moving along a straight line. The position of the particle at time t = 0 is x 0 = 5 m. (a) Find x for various times t by counting squares, and sketch x versus t. (b) Sketch the acceleration a versus t. (a) Count squares to t = 1 s, 2 s, 3 s, 4 s, 5 s, 6 s, 7 s, 8 s, 9 s, 10 s; add x(0) = 5 m (b) a = dv/dt = slope of v versus t curve. Take slopes at various times.
97* Figure 2-33 shows a plot of x versus t for a body moving along a straight line. Sketch rough graphs of v versus t and a versus t for this motion. Note: The curve of x versus t appears to be a sine curve; hence v(t) = dx/dt is a cosine curve and a(t) = dv/dt is a negative sine curve. 98 True or false: (a) The equation x = v 0 t + 1 2 at 2 is valid for all particle motion in one dimension. (b) If the velocity at a given instant is zero, the acceleration at that instant must also be zero. (c) The equation x = v av t holds for all motion in one dimension. (a) False; valid only for constant a. (b) False (c) True; by definition. 99 If an object is moving at constant acceleration in a straight line, its instantaneous velocity halfway through any time interval is (a) greater than its average velocity. (b) less than its average velocity. (c) equal to its average velocity. (d) half its average velocity. (e) twice its average velocity. (c) is correct, by definition of v av for constant acceleration. 100 On a graph showing position on the vertical axis and time on the horizontal axis, a straight line with a negative slope represents (a) a constant positive acceleration. (b) a constant negative acceleration. (c) zero velocity. (d) a constant positive velocity. (e) a constant negative velocity. (e) The slope represents the velocity; negative slope corresponds to a negative velocity