Sample Problem 2-2

Figure 2-6a is an x(t) plot for an elevator cab that is initially stationary, then moves upward (which we take to be the positive direction of x), and then stops. Plot v(t).

Fig. 2-6 (a) The x(t) curve for an elevator cab that moves upward along an x axis. (b) The v(t) curve for the cab. Note that it is the derivative of the x(t) curve (v = dx/dt). (c) The a(t) curve for the cab. It is the derivative of the v(t) curve (a = dv/dt). The stick figures along the bottom suggest how a passenger's body might feel during the accelerations.

Solution: The Key Idea here is that we can find the velocity at any time from the slope of the x(t) curve at that time. The slope of x(t), and so also the velocity, is zero in the intervals from 0 to 1 s and from 9 s on, so then the cab is stationary. During the interval bc, the slope is constant and nonzero, so then the cab moves with constant velocity. We calculate the slope of x(t) then as

The plus sign indicates that the cab is moving in the positive x direction. These intervals (where v = 0 and v = 4 m/s) are plotted in Figure 2-6b. In addition, as the cab initially begins to move and then later slows to a stop, v varies as indicated in the intervals 1 s to 3 s and 8 s to 9 s. Thus, Figure 2-6b is the required plot. (Figure 2-6c is considered in Section 2-6.)

Given a v(t) graph such as Figure 2-6b, we could “work backward” to produce the shape of the associated x(t) graph (Figure 2-6a). However, we would not know the actual values for x at various times, because the v(t) graph indicates only changes in x. To find such a change in x during any interval, we must, in the language of calculus, calculate the area “under the curve” on the v(t) graph for that interval. For example, during the interval 3 s to 8 s in which the cab has a velocity of 4.0 m/s, the change in x is

(This area is positive because the v(t) curve is above the t axis.) Figure 2-6a shows that x does indeed increase by 20 m in that interval. However, Figure 2-6b does not tell us the values of x at the beginning and end of the interval. For that, we need additional information, such as the value of x at some instant.

Sample Problem 2-3

The position of a particle moving on an x axis is given by

(2-5)

with x in meters and t in seconds. What is its velocity at t = 3.5 s? Is the velocity constant, or is it continuously changing?

Solution: For simplicity, the units have been omitted from Equation 2-5, but you can insert them if you like by changing the coefficients to 7.8 m, 9.2 m/s, and -2.1 m/s³. The Key Idea here is that velocity is the first derivative (with respect to time) of the position function x(t). Thus, we write

which becomes

(2-6)

At t = 3.5 s,

(Answer)

At t = 3.5 s, the particle is moving in the negative direction of x (note the minus sign) with a speed of 68 m/s. Since the quantity t appears in Equation 2-6, the velocity v depends on t and so is continuously changing.