2 Using the definitions of acceleration and velocity

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1 Physics I [P161] Spring 2008 Review for Quiz # 3 1 Main Ideas Two main ideas were introduced since the last quiz. 1. Using the definitions of acceleration and velocity to obtain equations of motion (chapter 2) 2. Adding and subtracting vectors via components (chapter 3) In chapter 4, these two ideas were then applied to various situations, like projectile motion and relative motion. 2 Using the definitions of acceleration and velocity Acceleration and velocity are defined as follows: a = d v dt v = d s dt In general, these are applied in one of two ways. 1. We are given an expression for position as a function of time and need to use the definitions to get expressions for velocity and acceleration 2. We are given an expression for acceleration as a function of time and need to use the definitions to get expressions for velocity and position Additional Note: An object s average velocity and average acceleration can be determined by applying the definitions to a finite period of time: a avg = v/ t and v avg = s/ t. Additional Note: For a constant acceleration, the average velocity will lie midway between the initial and final velocities. For a constant velocity, the average position will lie midway between the initial and final positions. 1

2 2.1 Example 1 Given the following expression for the position, s(t) = (2 m/s 3 )t 3 (3 m/s)t, find an expression for the velocity and acceleration. Since v is defined as d s/dt, take the derivative of s(t) to get the appropriate expression for v. Similarly, since a is defined as d v/dt, take the derivative of v(t) to get the appropriate expression for a. 2.2 Example 2 Given the following expression for the acceleration, a(t) = (2 m/s 3 )t, where the initial position and velocity are s(0) = 2 m and v(0) = 3 m/s, respectively, find an expression for the velocity and position. Since a is defined as d v/dt, that means that v = adt and so to find the appropriate expression for v we need to take the integral of a(t)dt. However, the integral is incomplete we need to also specify the initial velocity by adding v(0), which is 3 m/s in this case. The same procedure is used to find the expression for position. Additional note: For an object solely under the influence of the Earth s gravity (i.e., no air resistance, not being held) near the Earth s surface (i.e., not out in space), we find that its acceleration is 9.8 m/s 2 downward, which is indicated as g. If the object is placed on an inclined plane without friction, we find its acceleration is g sin θ, where θ, is the angle of the incline relative to horizontal. Thus, in those two cases, the problem might not need to explicitly state what the acceleration value is. 2.3 Example 3 An object undergoes an acceleration, a(t) = (2 m/s 3 )t, for 5 seconds and then undergoes an acceleration, a(t) = (2 m/s 2 ), for 10 seconds. If the initial position and velocity are s(0) = 2 m and v(0) = 3 m/s, respectively, how far has the object gone in the 15 seconds? In this case, the object experiences two different accelerations. You will need two sets of equations of motion, one for the first 5 seconds and one for 2

3 the last 10 seconds. Keep in mind that the initial position and velocity given in the problem statement are only valid for the first 5 seconds. You ll need the first set of equations to find the appropriate initial position and velocity values for the last 10 seconds. 3 Adding and subtracting vectors via components Vector quantities are those that have both a magnitude (represented by a number and unit) and a direction (usually given relative to some known direction). When we need to add or subtract vector quantities, it is easiest to first convert the vector quantities into component form (if they are not already given in that form), then add or subtract component by component. 3.1 Example 1 Given that A = (3 m)î + (2 m)ĵ and B = (2 m)î (3 m)ĵ, what is 2 A + B? Since the vector quantities are already in component form, we apply the expression 2A + B twice: once for the î components and again for the ĵ components. With the î components, we get 2(3 m) + (2 m) = 8 m. With the ĵ components, we get 2(2 m) + ( 3 m) = 1 m. The end result, then, is (8 m)î + (1 m)ĵ. 3.2 Example 2 Given that A = (3 m/s) at 43 and B = (5 m/s) at 143, what is 2 A + B? Since the vector quantities are not already in component form, you must first convert into component form. This is done in two steps. First, you must choose a coordinate system (sometimes the choice of coordinate system will greatly simplify the math you ll need to do). Second, use sine and cosine to determine the values of the components for each vector quantity. Once you have the vectors in component form, carry out the calculations separately for each component, as in the previous example. 3

4 Additional note: If the final answer is to be given in terms of magnitude and direction instead of component form, you must use the Pythagorean Theorem (C 2 = A 2 + B 2, for A and B perpendicular) and an inverse trigonometric function, like inverse tangent. Keep in mind that the inverse functions may not give you the appropriate direction you must interpret the direction given by the inverse function. 4 Projectile motion When dealing with problems involving vectors (like projectile motion), we really have two problems. The first has to do with identifying the relevant physics (as in chapter 2). The second has to do with solving the problem using vectors (chapter 3). To illustrate, I ll first examine the physics of projectile motion to obtain the relevant equations of motion. Then I ll explain how to use vector analysis to solve the problem. 4.1 Physics The physics of projectile motion is that an object is experiencing a constant downward acceleration with an initial velocity that is not downward. The acceleration need not be 9.8 m/s 2, as we can have projectile motion on other planets or moons. The basic physics, though, is that the acceleration is known to be a constant value, which means that the object s velocity will change in a steady, known way. If we indicate the acceleration as a then we can derive appropriate expressions for the velocity and position by applying the definitions of acceleration and velocity (as in chapter 2): Thus ends the physics. v = at + v(0) s = 0.5 at 2 + v(0)t + s(0) Additional Note: For objects solely under the influence of the Earth s gravity (i.e., no air resistance, not being held) near the Earth s surface (i.e., not out in space), the acceleration is known to be 9.8 m/s 2 downward. 4

5 4.2 Vector analysis This is a vector problem because we know the direction of the acceleration (down) while the object s velocity is not down rather, the velocity is changing toward down. As in chapter 3, we must first choose a coordinate system, split each vector value into the component values and then solve the relationships component by component. When choosing a coordinate system, it is easiest to choose one of the component directions to parallel to the acceleration direction. In this case, the acceleration direction is downward. Most people choose the two component directions to be horizontal (in the direction of motion) and upward. I ll do the same, indicating the two directions as ˆx and ŷ (you can use whatever letters you want). The three relationships identified in the physics portion of the problem (above) would be applied twice: once in each component direction. Additional Note: Rather than writing out each equation twice, you could use the ˆx and ŷ notation to keep track of the two component values within the same equation. Another common way is to write each vector value in matrix form. You can use whatever form you prefer. 4.3 Example A rock is thrown from an initial height of 1 m with an initial speed of 20 m/s at an angle of 45 degrees above the horizontal. How long does it take for the rock to reach its maximum height? Regardless of what is being asked for, we should first apply the relevant physics and identify the appropriate equations of motion. We can find the requested piece of information later. As with all projectile motion problems, the three relevant equations of motion are identified above. Since we ve identified our component directions, though, the vector values of the acceleration, initial velocity and initial position should be written in component form: a = (0)ˆx + ( 9.8 m/s 2 )ŷ v(0) = (14.1 m/s)ˆx + (14.1 m/s)ŷ 5

6 s(0) = (0)ˆx + (1 m)ŷ Notice that I set the initial position to be (1 m)ŷ, which means I ve set the ground to be a position of zero. If I look only at the ˆx components, I have the following three relationships: a x = 0 v x = 14.1 m/s s x = (14.1 m/s)t whereas for the ŷ components, I have the following three relationships: a y = ( 9.8 m/s 2 ) v y = ( 9.8 m/s 2 )t + (14.1 m/s) s y = ( 4.9 m/s 2 )t 2 + (14.1 m/s)t + (1 m) At this point, we look for what is being asked for. In this case, it is asking for the time when the maximum height is achieved. The maximum height is achieved when the rock is neither moving up or down (i.e., v y is zero). Plug in zero for v y and solve for t. 5 Relative motion As with projectile motion, when dealing with problems involving relative motion in more than one dimension, we really have two problems. The first has to do with identifying the relevant physics. The second has to do with solving the problem using vectors. 5.1 Physics The physics of relative motion is that an object s velocity is always measured relative to some other object. In some cases, it is useful to measure an object s velocity relative to two different other objects. In those cases, we need a way of relating the various relative velocities. The basic relationship is as follows: v 1 rel 3 = v 1 rel 2 + v 2 rel 3 6

7 In other words, the velocity of object 1 relative to object 3 is the sum of the velocities using object 2 as an intermediate object. In these cases, then, it is important to identify the three objects and how the three velocities are related prior to doing any math. 5.2 Vector analysis When two of the relative velocities are known and those two are not paralle, this becomes a vector problem. As in chapter 3, we must first choose a coordinate system, split each vector value into the component values and then solve the relationship component by component. 5.3 Example I am on a train that is moving at 20 m/s in a direction 23 east of north relative to the ground. If I roll a ball across the train at a velocity of 5 m/s in a direction 43 west of north relative to the train, what is the velocity of the ball relative to the ground? Regardless of what is being asked for, we should first apply the relevant physics and identify the appropriate relationship(s). We can find the requested piece of information later. As with all relative motion problems, the basic relationship is identified above. The three objects in this case are the train, the ball and the ground. In this case, the train is the intermediate object. v ball rel ground = v ball rel train + v train rel ground For the component directions, let s choose north and east, indicated as ŷ and ˆx. Since we ve identified our component directions, we can now write the vector values in component form: v ball rel train = ( 3.41 m/s)ˆx + (3.66 m/s)ŷ v train rel ground = (7.81 m/s)ˆx + (18.41 m/s)ŷ Plugging in, I have the following relationships: v x,ball rel ground = ( 3.41 m/s) + (7.81 m/s) v y,ball rel ground = (3.66 m/s) + (18.41 m/s) 7

8 This gives me the velocity of the ball in terms of its two component values. To find the magnitude and direction of the ball, I use the Pythagorean Theorem and an inverse trig function (like tan 1 ). 6 One final word Every problem involves first identifying the physics. After all, it is the physics that is important, not the individual problem. So, expect to be presented with a problem that you ve never seen before. Don t panic. The first thing is to identify the physics. 8