Physics 201, Fall, 2010, Week Two

The class lectures will include most- but perhaps not all- of the points covered in class. I also include fully worked out problems (FWOPs) as examples to help you. In week two, we cover motion in one dimension.

Displacement Since we are discussion moving in one dimension, all we need is to distinguish the one direction and its opposite, so we will use (+) for along the direction (call it z-direction) and (-) for opposite that direction. The unit we use is the unit of length, “meter.”

In this class, we only consider three types of displacement along the z-direction: constant location (no speed or acceleration), constant speed, and constant acceleration. You standing still is a simple example of constant location. Driving a car with cruise control is close to a realistic example of constant speed. Throwing a ball up, or down, near the Earth’s surface is a region where the acceleration of gravity is very close to constant.

Velocity There is a difference between the words “speed” and “velocity”: speed refers to how fast you are going, while velocity refers to both how fast and in what direction you are going. So if the one dimension- one direction- I am going in is from left to right, and I am going at a speed of 20 miles/hour, I need to say which direction (left or right) I am going in to specify the velocity.

The velocity can be described another way: the change in location with time. The full answer of how location and velocity connect involves calculus, but we can get the idea of an “average” velocity by saying:

Average velocity = [change in location]/ [change in time] (Eq. 2.1)

Let me go through a few examples of this. Example 2.1: Suppose I am running from left to right at a constant speed and I cover a distance of 50.0 meter in 8.0 second. What is the velocity? The speed = 50.0/8.0 = 6.3 m/s. To distinguish the left-to-right running from the right-to-left running, there are several ways. I’ll choose to call left-to-right running (+), and right-to-left running (-). Then the velocity = + 6.3 m/s.

Example 2.2: Suppose I run a distance of 50.0 meter in 8.0 second, then I stop to rest for 12.0 second, then I run another 50.0 meter in a further 8.0 second. What is the average velocity? The average speed = distance moved/ elapsed time = 100.0 m/28.0 s = 3.57 m/s. The person is running left-to-right, so the velocity = + 3.57 m/s. 2

Example 2.3: Now I run 50.0 meters in 8.0 seconds, then I rest for 12.0 second, then I run back 50.0 m. to the original location in a further 8.0 seconds. What is the average velocity? The elapsed distance = 0, so both the speed and the velocity are zero. Notice! Even though the person (the object) moved, there was no average change in distance (location), so the average speed and velocity are zero.

Acceleration

The word acceleration has multiple meanings in our language & culture. In science & engineering, it has a limited, specific meaning:

Acceleration = [change in velocity]/[change in time] (Eq. 2.2)

Since velocity has both a direction and a size, acceleration likewise has a direction and size. We will be working only with the average acceleration idea, not the acceleration at an instant of time. Why? Because we need calculus to figure out the instantaneous acceleration, and our class does not use calculus.

Another reason for including acceleration, but not the change of acceleration with time, is historical: the gravitational acceleration near the Earth’s surface is virtually constant, so there is no change in this acceleration with time. In fact, many of our examples use this acceleration, which is 9.81 m/s2 directed downward.

Having introduced the definition and basic idea, let’s look at some examples to see how the idea of acceleration works in practice.

Example 2.4: Suppose a car starts from rest, moving from right-to-left, and accelerates at a constant value, reaching 45 m/s in 3.7 seconds. What is the acceleration? Notice! We are asking for both a size and a direction when we ask for an acceleration. The direction, by our convention, is negative (-). What of the size? Acceleration = change in velocity/change in time = -(45 m/s) / (3.7 s) = - 12.2 m/s2.

Example 2.5: Suppose that I drop a ball from rest and it falls under the influence of gravity for 4.8 seconds. How fast is it going after this much time? Acceleration = change in velocity/ change in time = 9.81 m/s2 = (change in velocity)/(4.8 s) So: change in velocity = 47 m/s downward. 3

Next, let’s look into the situation where the acceleration due to gravity acts on an object. Suppose I throw an object straight up into the air. What happens? The object goes upward, slowing down, and eventually stops. It then begins to fall, moving faster and faster downward. All throughout the motion, the acceleration is the same. The acceleration 2 is constant, and is 9.81 m/s downward. Since the acceleration is constant, the final velocity (vF ) and the initial velocity (vI ) are related by: vF – vI = (-9.81) (elapsed time) (Eq. 2.3)

Now, since the velocity is changing at a constant rate, the average velocity is exactly half-way between vF and vI. So the average velocity is: vAVG = (vI ) + (1/2) (vF – vI ) = (vF + vI )/2 (Eq. 2.4)

What distance has the object moved due to its initial velocity and acceleration? The distance moved and the average velocity are related: distance moved = (vAVG ) (elapsed time) = [(vI ) + (1/2) (vF – vI )] (elapsed time) = [(vI ) + (1/2) (acceleration) (elapsed time)] (elapsed time) 2 = (vI ) (t) + [(1/2) (a) (t) ] (Eq. 2.5)

Let’s go through some examples to illustrate some of the consequences of these ideas & equations.

Example 2.6: Suppose I throw a ball upward with an initial speed of 15 m/s. What speed does it have when it returns to the same vertical distance? Here the distance moved is zero, so:

0 = (15) (t) + [(9.81/2) t2] = [15 + 4.9 t] (t)

This has two solutions: (1) t = 0, that is, the initial time, and (2) t = 15/4.9 s = 3.1 s. For solution (1), the speed is 15 m/s upward. For solution (2): vF – vI = (9.81) (elapsed time) ----> vF = (+ 15 m/s) – [(9.81) (15/4.9)] = - 15 m/s.

So the ball has the same speed, but opposite direction, when it returns to the initial vertical location. 4

Example 2.7: An object having a mass of (5.0 kg) moves initially upward at a speed of (22 m/s) against gravity. How long does it take to reach its apogee? How far above the initial location is the apogee?

The equations we have available include:

Equation 2.3: vF – vI = (-9.81) (elapsed time) What does this tell us? In this instance, the final velocity (at the apogee) is zero, so the elapsed time = (initial speed)/ (9.81) = 22/ 9.81 = 2.2 seconds.

2 2 Equation 2.5: distance moved = (vI ) (t) + [(1/2) (a) (t) ] = 22t + (-4.9 t ). For a time of 2.2 seconds, the distance moved = 48.4 m. + 23.7 m. = 25 m. Notice, also, that these numbers are independent of the mass of the object- the acceleration due to gravity, and the resulting speeds and distances, are the same for all objects as long as we are near the surface of the Earth. When we get far enough away from the surface for the finite size of the Earth to matter, we will find a qualitatively different answer.

One dimensional motion with constant acceleration

This is the situation near the Earth’s surface, and the examples we have discussed above include several where there is a constant acceleration. The key idea is: distance moved = (vAVG ) (elapsed time) = [(vI ) + (1/2) (vF – vI )] (elapsed time) = [(vI ) + (1/2) (acceleration) (elapsed time)] (elapsed time) 2 = (vI ) (t) + [(1/2) (a) (t) ] (Eq. 2.5)

Example 2.8: So let us compare some situations. (a) Suppose that two balls are both 40.0 m above a flat, horizontal surface. Ball A is thrown downward with an initial speed of 12.0 m/s. Ball B is thrown upward with the same initial speed. Which ball will hit the ground first? This is easy- ball A, of course. How soon after ball A hits the ground will ball B hit the ground?

To answer this question, we need the time each ball is in the air. Let’s apply Eq. 2.5. Ball A: distance moved = -40.0 m (negative since the distance is down, not up) = (-12.0 m/s) (t) – 4.9 t2 which leads to: t = 1.9 s.

Ball B: distance moved = - 40.0 m = (+12.0 m/s) (t) – 4.9 t2 , which leads to t= 4.3 s.

Thus, ball B hits the ground 2.4 s. after ball A does so. 5

(b) We have two balls, A and B. Ball A is initially 120.0 m. above a flat, horizontal surface. Ball B is initially 45.0 m. above the same surface. Ball A falls (downward) from rest. Ball B is thrown upward with an initial speed of 25.0 m/s. How soon after they start moving, and at what location above the surface, do they hit (or pass) each other?

How do I start? I realize that the time in the air before the balls hit, or pass, each other is the same, and I equate these two times. Ball A time:

This is based on equation 2.5, with ball A starting 120.0 m above the surface. Thus, at any time later, the location of ball A above the surface is:

(120.0 m.) + distance moved = 120.0 m. + (-4.9 t2 )

What about ball B? We use the same idea, except that ball B has a different initial location above the surface and a non-zero initial speed. The location of ball B above the surface at the same time is:

(45.0 m.) + distance moved = 45.0 m. + [(25.0 m/s) (t)] + [-4.9 t2 ]

Before we go any further, please notice something important: gravity- the acceleration- has the same effect on both balls. So now we find a time (t) at which these two locations are the same:

120.0 m. + (-4.9 t2 ) = 45.0 m. + [(25.0 m/s) (t)] + [-4.9 t2 ]

The (t2 ) term due to the acceleration cancels, leaving:

75.0 m. = (25.0 m/s) (t) or: t = 3.00 seconds is the time at which the balls hit or pass. The location above the surface at this time is:

120.0 m. + (-4.9 t2 ) = 120.0 – 44.1 = 75.9 m.

Freely falling objects

Throughout this chapter, we have been discussing the various effects of freely falling objects. We have used the acceleration due to gravity, 9.81 m/s2 downward. However, the methods are the same regardless of the size and/or the direction of the acceleration. The restriction in this entire chapter is that we are only looking at motion in one dimension- one direction. In subsequent chapters, indeed starting with chapter three, we will discuss situations in which there are different accelerations in different directions. For example, in our everyday experience, we have an acceleration due to gravity in one direction, and no acceleration in the other two directions. Because we use coordinate systems in which the coordinates are perpendicular, the 6 effect of gravity is only in one dimension, and the other two have objects moving with no change of velocity.

2.1 A particle is moving along the x-axis. Its position is given by the equation: x = 2 + 3t – 4t2 where (x) is in meters and (t) is in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t = 0.

Answer: At t = 0, x = 2 m. So I set the left hand side equal to 2 = 2 + 3t – 4t2 ----> 0 = (3-4t) (t) Which has two solutions, t = 0 and t = (3/4 s). In part (a) I am to find the position when the particle changes direction. How can I find out when and where the particle changes direction. Initially, the particle is moving to the right (+ direction). When does it change direction? When the speed is zero. What is the speed?

Speed = change in direction/change in time = {[(2 + 3(t + Δt) – 4(t + Δt)2 ] – [2 +3t – 4t2 ]}/(Δt) = 3 – 8t = 0 when t = 3/8 s. What is the position at this time? x = 2 + 3t – 4t2 = 2 + [3(3/8)] – [4(3/8)2] = 41/16 m.

For (b), I use the time (3/4 s) at which the object returns to its initial position. At this time, the speed = 3 – 8t = - 3 m/s.

2.2 A driver moving at 30.0 m/s enters a one-lane tunnel and sees a slow-moving van 155 m. ahead, moving at 5.00 m/s. The driver brakes, but due to the wet road can only accelerate at -2.00 m/s2 . Will there be a collision? If so, determine how far into the tunnel and at what time the collision occurs. If not, determine the distance of closest approach between the vehicles.

Where do I start? I begin by expressing how each vehicle changes position with time. I use x = 0 as the location of the driver at t = 0, which means at t = 0, the van is at x = + 155m. The equations for the car and van are:

Car: x(t) = [(30.0) t] – [(1/2) (2) t2 ] Van: x(t) = [+ 155m.] + [(5.00) t]

If there is a collision, the two objects have the same distance at some time:

[(30.0) t] – [(1/2) (2) t2 ] = [+ 155m.] + [(5.00) t]

Which can be rewritten as:

(t2 ) – (25.0 t) – 155 = 0

This has two solutions: t = 11.4 s and t = 13.7 s. 7

At t = 11.4 s, both vehicles are at x= 212 m. Further, the speed of the car (7.2 m/s) is still greater than the speed of the van (5.00 m/s) [CAN YOU EXPLAIN HOW I GOT THESE NUMBERS?], so there is a collision at this time and place.

2.3 A person throws a set of keys vertically upward to her sister, who is in a window 4.00 m. above. The sister catches the keys 1.50 s. later. a) What was the initial speed of the keys? b) What was the speed of the keys just before the sister caught them? Answer: (a) As with earlier examples, this is an application of equation 2.5:

2 2 distance moved = + 4.00 m = (vI ) (t) + [(1/2) (a) (t) ] = (vI ) (1.50) – [(4.9) (1.5) ] = (vI ) (1.50) – 11.0 ------> (vI ) = 10.0 m/s.

(b) This is an application of equation 2.3: vF – vI = (-9.81) (elapsed time) = - 14.8 m/s -----> vF = -4.8 m/s is the speed

What is the physical significance of the negative sign? It turns out there are two times at which the keys are at the 4.00 m. height of the window. One is 0.55 s, when the keys are going up, and the other is 1.50 s, when the keys are coming down. The sister catches the keys when they are coming down.