Scalar versus Vector Quantities Scalar Quantities Magnitude (size) 55 mph Speed Average Speed = distance (in meters) time (in seconds) Vector Quantities Magnitude (size) Direction 55 mph, North v = Dx Dt Speed: Example Two During a 3.00 second time interval, a runner s position changes from x 1 = 50.0 m to x 2 =30.5 m. What was her average speed? (Ans: -6.50 m/s) A car is driven at an average speed of 100.0 km/hr for two hours, then driven at an average speed of 50.0 km/hr for the next hour. What was the average speed for the three-hour trip? (Ans: 83.3 km/hr) Remember that average speed is calculated using: v=x/t Since the speed changed, we have to address each part of the trip separately. 1. To find the distance for the first two hours, we use: x=vt x= 100km/hr X 2 hr = 200 km 3. To find the average velocity, we will divide the total distance by the time: v=x/t = 250 km /3 hr = 83.3 km/hr 2. To find the distance for the last hour, we use the same process: x= 50km/hr X 1 hr = 50 km 1
Average vs. Instantaneous Speed Average Speed The speed averaged for an entire trip Instantaneous Speed This is the speed at an infinitely short interval (a speedometer reading taken at any instant) Constant velocity Changing velocity Acceleration Acceleration A change in velocity Can be a change in speed (speeding up or slowing down) Acceleration (m/s 2 ) a = Dv = v final v initial Dt time Speeding up or slowing down Can be a change in direction (turning a bicycle) Acceleration can be negative (slowing down, decelerating) Superman is flying through space and slows down from 24,000 m/s to 20,000 m/s in 5 seconds. What is his acceleration? What is unusual about this number? A car starts from a complete stop at a stoplight and accelerates at 5 m/s 2 for 10 seconds. What is its final velocity? (Ans = 50 m/s) 2
Kinematic Equations A car moving at 30 m/s decelerates at 3.0 m/s 2. How long will it take for it to come to a complete stop? v = v o + at x = x o + v o t + ½ at 2 (ANS: t = 10 seconds) Timeindependent a = v 2 - v o 2 2x v ave = v + v o 2 An airplane takes off at 27.8 m/s and can accelerate at 2.00 m/s 2. How long must the runway be if the plane is to take off safely? How long does it take a car to cross a 30.0 m intersection if it accelerates at 2.00 m/s 2? (Ans: 193 m) (Ans: 5.48 s) A pitcher throws a baseball at 44 m/s. What is the acceleration if the ball travels through 3.5 m from the start of the pitch to the release? You wish to design an airbag that can protect a driver from a head-on collision at 100 km/h. Assume the car crumples on impact a distance of 1.00 m. a) What is the driver s acceleration? b) How fast does the airbag need to act? (Ans: 280 m/s 2 ) 3
A driver is moving at 28 m/s. It takes him 0.50 seconds to react and slam on the brakes. At that time, the brakes provide an acceleration of 6.0 m/s 2. Calculate the total stopping distance. Graphing Given the following x-t graph a. Sketch a v-t graph b. Sketch an a-t graph (Ans: 79 m) Graphing Given the following x-t graph a. Sketch a v-t graph b. Sketch an a-t graph (assume a 0.25 s turnaround) 4
Graphing Given the following v-t graph: a. Sketch an x-t graph b. Sketch a v-t graph 5
Given the following Velocity versus Time graph: a. Sketch an Acceleration versus Time graph b. Draw a Displacement versus Time graph. a. Given the following x-t graph, draw a v-t graph. b. Describe the car s motion 6
The following graph is for the motion of an elevator a. When does the elevator have the least speed? b. When is it moving the fastest? c. Sketch a v-t graph Link to Calculus Displacement Slope Area (Derivative) (Integrate) Velocity Slope Area (Derivative) (Integrate) Acceleration If given the velocity function x= position x = x o + v o t + ½ at 2 Dx = dx = v v = v o + at Dt dt Dv = dv = a a = v final v initial Dt dt t dx = v(t) dt dx = vdt xo x dx = 0t vdt x x o = 0t vdt If given the acceleration function dv = a(t) dt dv = adt The function for the position of a particle is s = 2t 2 m. a. Determine the function for the velocity. b. Calculate the velocity at 4 seconds c. Determine the function for acceleration. vo v dv = 0t adt v v o = 0t adt 7
A particle starts with an initial velocity of 10 m/s. a. Calculate the velocity at 8 s b. Calculate the velocity at 10 s The speed of an automobile (m/s) is given by the following equation: v = 4 + 2t 3 a. Find the equation for position. b. Calculate how far has the car travelled between t=0 and t = 2s. (16 m) c. Find the equation for acceleration. d. Calculate the acceleration at 2s. (24 m/s 2 ) Calculus: Ex 2 Find the position and velocity equations of an object that has an acceleration of a(t) = 2t 4, an initial velocity of +4 m/s and starts at x=0. v v o = 0t adt v v o = 0t (2t-4)dt v v o = t 2 4t v 4 = t 2 4t v = t 2 4t + 4 x x o = 0t vdt x x o = 0t (t 2 4t + 4)dt x x o = 1/3t 3 2t 2 + 4t Calculus: Ex 2 Determine the equations for the velocity and acceleration of a particle whose position is given by: x = 9.75 + 1.50t 3 v = dx/dt = 4.50t 2 a = dv/dt = 9t Calculate the average velocity between t = 2s and t =3s v = Dx/Dt x 2 = 9.75 + 1.50t 3 = 21.75 m x 3 = 9.75 + 1.50t 3 = 50.25 m v = (50.25 m 21.75m) = 28.5 m/s (3s 2s) 8
Calculate the instantaneous velocity and acceleration at 2s. v = 4.50t 2 = 18 m/s a = 9t = 18 m/s 2 A box is being pulled a student and its position varies by: x=0.5t 3 + 2t. What is the speed of the box at t = 0 s and at t = 2s? v = dx/dt v = 1.5t 2 + 2 v(0) = 2 m/s V(2) = 8 m/s You are given the following v-t graph. A. Calculate the distance travelled in the first three seconds. B. Calculate the function for the line C. Determine the formula for the position through integration. Given the following v=t graph: a. Where is the turning point b. Calculate the formula for velocity c. Calculate the distance travelled from 0 to 6s d. Determine the formula for position e. Determine the formula for acceleration Falling Objects Aristotle (Greece, ~400 B.C.) Heavier object fall faster than lighter ones Galileo (1600 s, Renaissance) All objects fall at the same rate (near the surface of the Earth, in a vacuum) Aristotle versus Galileo. In the steel cage. 9
Distance (m) Graph Galileo s Numbers Determine the formula for distance fallen from: y = y o + v o t + 1/2at 2 Distance vs. Time 60 50 40 30 20 10 0 0 0.5 1 1.5 2 2.5 3 Time (s) A rock is dropped into a well and hits the water in 2.45 seconds. a. How deep is the well? (29.4 m) b. Calculate the impact velocity (24 m/s) A ball is thrown off a cliff with a initial velocity of 3.00 m/s. a. Calculate its position and speed after 1.00 s. (7.90 m, 12.8 m/s ) b. Calculate its position and speed after 2.00 s. (25.6 m, 22.6 m/s) 10
You throw a ball into the air with an initial velocity of 15.0 m/s. a. Calculate the maximum height of the ball. (11.5 m) b. Calculate the total time the ball was in the air. (3.06 s) c. Calculate the velocity when it returns to your hand. d. What time will the ball pass a point 8.00 M above the person s hand? (0.69 s and 2.37 s) You throw a ball into the air with an initial velocity of 27.5 m/s. a. Calculate the maximum height of the ball. (38.6 m) b. Calculate the total time the ball was in the air. (5.6 s) c. Calculate the velocity when it returns to your hand (27.5 m/s). d. What time will the ball pass a point 10.0 M above the person s hand? (0.39 s and 5.22 s) Inclined Plane a s = + gsinq q a g q a. A car travels down a hill with a 8.00 o slope. Calculate the acceleration. b. Calculate the velocity at the bottom of a 150.0 m hill. A skier travels down a 100-m long slope. At the bottom, her speed is 20 m/s. Calculate the angle of the slope. (a = 2 m/s 2, q = 12 o ) 11
Acceleration (m/s2) Displacement (m) Velocity (m/s) Velocity vs. Time Given the following Velocity versus Time graph: a. Sketch an Acceleration versus Time graph (slope) b. Draw a Displacement versus Time graph. (area) 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 0 1 2 3 4 5 6 7 8 9 10 11 Time (s) Displacement vs. Time Acceleration vs. Time 5 4 3 2 1 0 0 2 4 6 8 10 12 Time (s) 250 200 150 100 50 0 0-50 2 4 6 8 10 12 Time (s) x = t 2 Chapter One 23.a) 9.12 X 10-6 s b) 3420 m c) 440 m/s d) 22 m/s 24. a) 0.20 m b) 20 m/s c) 27 m/s d) 9.0 X 10-3 m 2 25. a) 3600 s b) 8.64 X 10 4 s c) 3.16 X 10 7 s d) 9,75 m/s 2 26.a) 7.0 m b) 1.0 X 10 5 m c) 30 m/s d) 0.16 m 27.a) 12 in b) 50 mph c) 3 miles d) 1/4 in 2.a) -200 yd/min, +333 yd/min b) +120 yd/min 4. a) 48 mph b) 50 mph 6. Turning point at t = 3s 8. Zero acceleration 0 to 2s, linear increase 2-4s 10. a) 6 m b) 4 m/s c) 2 m/s 2 12. a) 8.75 m/s 2 b) g 14. t P = 15.1 s t H = 15.3 22. a) 1.67 m b) 2 m/s c) 4 m/s 2 26. -10m/s, -20 m/s, 75 m/s 28. 0 m/s, 5 m/s, 20 m/s, 30 m/s, 30 m/s 12
16. Graph with slope -9.8, crosses x-axis at 2 s 18. a) v = -24 m/s b) 4.5 s 20. x = 134 m 46. a) 100 m b) -2 m/s 2 c) 11 s 52. a) 12.5 m b) 45 s 13