InterviewSolution
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InterviewSolution
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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). |
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Answer» Solution :KEY IDEA We can find the velocity at any timefrom the slope of the `x(t)` curve at that time. Calculations : 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 `(Delta x)/(Delta t) = v = ( 24 m - 4.0 m)/(8.0 s - 3.0 s)` ` = + 4.0 m//s` (2-5) 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 Fig. 2-6 b. In addition, as the cab Figure 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 = d x // dt`). (c) The `a (t)` curve for the cab . It is the derivativeof the `v(t)` curve ( a = dv/dt). The stick FIGURES along the bottom suggest how a passenger.s body might feel during the accelerations. initially beging to move and then later slows to a stop, v varies as indicated in the intervals 1 s to 3s and8 s to 9s. THUS, Fig. 2-6 b is the required PLOT. ( Figure 2-6c is considered in Section 2.6 ) Given a v(t) graph such as Fig. 2-6 b, we could " work backward" to produce the shape of the associated `x(t)` graph (Fig. 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 the that interval. For example, during the interval 3s to 8 s in which the cab has a velocityof 4.0 m/s. the change in `x` is `Delta x = ( 4.0 m//s) (8.0 s - 3.0 s) = + 20 m`. (2-6) (The 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, Fig. 2-6 b 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. |
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