(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is 115 m/s at t=20 s. (b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is 5.0 m/s2 .
Solution:
Part A
Figure A shows the approximate slope of the curve at time 20 seconds.
To solve for the slope of this line, we need to approximate by using two points. In this case, we shall use the points at time 15 seconds and 25 seconds.
Approximately, when t=15\ \text{s}, the position is x=1000\ \text{m}, and when t=25\ \text{s}, the position is x=2150\ \text{m}. Thefore,
\begin{align*}
\text{velocity }& = \text{slope} \\
v& = \frac{\Delta x}{\Delta t} \\
v& = \frac{x_2-x_1}{t_2-t_1} \\
v& = \frac{2150\ \text{m}-1000\ \text{m}}{25\ \text{s}-15\ \text{s}}\\
v& = 115\ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Part B
One can immediately figure out from the given graph that it is a straight line. The slope of the line can be computed by using any two points in the line.
Here, v=15\ \text{m/s} when t=0\ \text{s}, and v=40 \ \text{m/s} when t=5\ \text{s}. The acceleration is
\begin{align*}
\text{acceleration}& = \text{slope} \\
a& = \frac{\Delta v}{\Delta t}\\
a& = \frac{v_2-v_1}{t_2-t_1} \\
a& = \frac{40\ \text{m/s}-15\ \text{m/s}}{5\ \text{s}-0\ \text{s}}\\
a& = 5.0\ \text{m/s}^2 \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
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