2-4	Acceleration and Graphs of Accelerated Motion

We have looked at several situations in which an object's velocity changes over time. We know that an object, in everyday language, can pick up or lose speed. A powerful sports car picks up speed more quickly than an economy car; a braking car loses speed more quickly on dry pavement than on an icy road. It is therefore useful to talk about the rate at which it happens. But we want to do so in language consistent with describing motion in terms of position, displacement, and velocity. We therefore will define acceleration as the rate at which velocity changes with time. If an object's velocity is v₁ at instant t₁ and v₂ at instant t₂, then we define average acceleration as follows:

	Definition: Over the interval Δt = t2 − t1, (2-7) for one-dimensional motion.

Procedure  2-2
Determining Average Acceleration in One Dimension
	1.   Determine an object's instantaneous velocity v1 at time t1. 2.   Determine its instantaneous velocity v2 at later time t2. 3.   Calculate the increments Δv = v2 − v1 and Δt = t2 − t1. 4.   Use Equation 2-7 to calculate a .

Let's see how this definition works in the case of the cart in Figure 2-13a. The cart leaves the hand in the downhill direction at t = 0 , but soon reverses direction because of the suspended weight. Look at its v versus t graph (Figure 2-12a). At t₁ its velocity is negative; it is moving to the left. At t₂ its velocity is zero. Thus, over the interval Δt = t₂ − t₁,

By similar arguments, we can conclude that the average acceleration of the cart in Figure 2-13b (which goes right immediately after leaving the hand at t = 0, but later goes left) is negative throughout. We can now summarize our reasoning:

Time Interval Δt Change in Velocity Δv What Object Is Doing Cart in Figure 2-13a t2 − t1 = +value + Moving left but slowing down   t3 − t2 = +value + Moving right and speeding up Cart in Figure 2-13b t2 − t1 = +value − Moving right but slowing down   t3 − t2 = +value − Moving left and speeding up

 Units of Average Acceleration Average acceleration is the rate of change of velocity. For instance, if a car went from 0 to 60 mi/hr in 5 seconds, it would be gaining an average of 12 mi/hr each second, or per second. Its acceleration in mixed units would be 12 (mi/hr)/s In SI, the units of average acceleration are m/s (meters per second) divided by s, giving us (m/s)/s (meters per second per second). If, for example, your velocity changes from 15 m/s to 35 m/s as your stopwatch advances from 3 s to 7 s, your velocity increases on the average by 5 m/s during each second of the 4-s time interval. Then

 Average Acceleration as Slope of v Versus t Graph If we apply the general definition of slope

	The slope of a v versus t graph between t1 and t2 is equal to the average acceleration over that interval.

The graph in Figure 2-12a rises left to right—its slope is always positive. The descending graph in Figure 2-12b always has a negative slope. But the slope is the average acceleration. So as we concluded previously, the acceleration is always positive for the motion graphed in Figure 2-12a and always negative for the motion graphed in Figure 2-12b.

 Instantaneous Acceleration Like average velocity, average acceleration is defined over an interval. As we did for average velocity in WebLink 2-1, we can find average accelerations for progressively smaller time intervals. The value approached as the interval closes in on a particular instant is defined to be the acceleration at that instant.

	Definition: In one dimension, (2-8) Without a bar, a denotes instantaneous acceleration.

When the v versus t graph is not a straight line, we can “zoom in” on a segment that includes any instant t that interests us. If we zoom in close enough, the segment is essentially straight, and its slope gives the instantaneous acceleration at that instant.

	Example 2-5		Determining Instantaneous Accelerations by Finding Slopes
	The motion of a certain object is represented by the v versus t graph in Figure 2-14. Find approximate values of the instantaneous acceleration a.   at t = 2.0 s b.   at t = 20.0 s Figure 2-14  Graph of v versus t for Example 2-5. Solution Choice of approach. Pick a small segment of the graph that includes each instant, then find the slope of each segment. The mathematical solution. a.   To include t = 2.0 s, we arbitrarily pick the interval from t1 = 1.5 s to t2 = 2.5 s. (You can redo this choosing an even smaller interval to see how much it changes your result—see Problem 2-100 at the end of the chapter.) From the graph, we find that the velocity at t1 = 1.5 s has the value v1 = 10.2 m/s, and at t2 = 2.5 s, its value is v2 = 15.6 m/s. The slope of the chosen segment is Then at t = 2.0 s, the instantaneous acceleration a ≈ 5.4 (m/s)/s or 5.4 m/s2 . b.   As we go beyond t = 10 s, we see the graph becoming more nearly horizontal. Because the slope of any horizontal line is zero, the slope of a small enough segment including t = 20.0 s is approximately zero, and so a ≈ 0. Making sense of the results. The value obtained in b reflects the fact that by t = 20 s, the velocity is no longer changing.  Related homework: Problems 2-28, 2-29, and 2-100.

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