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Now we have three quantities, position, velocity, and acceleration, all related to each other algebraically. Often it is helpful to visualize these quantities graphically. The following principles apply
  1. Given a graph of x versus t, the instantaneous slope at time t is the velocity v at time t.
  2. Given a graph of v versus t, the instantaneous slope at time t is the acceleration a at time t.
  3. Given a graph of a versus t, the area under the curve during interval Δt gives the change in velocity v during that interval.
  4. Given a graph of v versus t, the area under the curve during interval Δt gives the change in position x during that interval.

The graph of a versus t for a car which undergoes constant acceleration is shown in Figure 2-7. Sketch the graph of v versus t. Assume v = 0 m/s at t = 0 s.


The area under the curve between 0 and Δt is shown in a "forward-slash" hatch. This area is Δv, that is, the change in velocity during Δt. The reason for principle 3 above becomes clear if we recall the formula for the area of the rectangle representing the hatched region:


area = height x length
Δv = aΔt


This is how we defined acceleration in equation (4). During the second interval Δt, the area is a Δt again. Thus the change in velocity is the same, as shown in Figure 2-8. For the next intervals of time, the quantity Δv is constant.

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Figure 2-7
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Figure 2-8

Note that
Figures 2-7 and 2-8 give (almost) the same information in different forms.
Figure 2-7 has the information that the acceleration is positive and constant, so the car is speeding up (if it is going forward).
Figure 2-8 has the information that the velocity is increasing at a constant rate.
This is the same thing.


Before you read the next example, consider an object thrown straight up. When it reaches the top of its path, what is the direction of its velocity? What is the direction of its acceleration?

An apple is tossed straight up in the air. The graph of y versus t is shown in Figure 2-9. Sketch the graphs of v versus t and a versus t.


To obtain an instantaneous slope, we can use an imaginary electron microscope to look at a small portion of the graph. A small section of Figure 2-9 has been enlarged using such a microscope. This portion looks almost straight, so we could calculate its slope if we had some numbers. We can at least read that the slope is positive and very large, hence the first point in Figure 2-10.

..\art 2 jpg\figure 2-ti.jpg ..\art 2 jpg\figure 2-tj.jpg
Figure 2-9 Figure 2-10
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Figure 2-11 Figure 2-12

The second point on Figure 2-10 still has a positive slope, but smaller. The third point has a zero slope (see uppermost point in Figure 2-9). The fourth point has negative slope, and the fifth point has a slope more negative still.

It will not come as a surprise if we draw a straight line through these points, as in Figure 2-11. We take the slope at three points, but it is easy to see that the slope is constant and negative. We graph the acceleration in Figure 2-12.

Does this match your expectation? Particularly at the top of flight, did you know that the direction of the acceleration would be down?



Figure 2-13 shows v versus t for a car. Sketch the graphs for x versus t and for a versus t. (Say x = 0 at t = 0.)


Let's graph x versus t first. From t = 0 to 1 s, there is no area under the curve, so x stays constant. From t = 1 to 2 s, the area under the curve is 0.5 m. (Recall the area of a triangle is A = 1/2 base x height.)

From t = 2 to 3 s the area under the curve is 1.5 m, so that the x value jumps to 2 m (see the first graph of Figure 2-14). Between t = 3 and 4 s, the area is 2 m and x jumps to 4 m, and so on. Figure 2-14 shows the result.

For the graph of a versus t, the slope of v versus t for any point between t = 0 and 1 s is zero. The slope jumps to 1 m/s2 for the interval from 1 to 3 s and drops back down to zero for times after 3 s. (See Figure 2-15.)

Think about all three graphs for a while and note how they give the same information in different forms.

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Figure 2-13 Figure 2-14
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Figure 2-15



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