Question 3.1: A car moving along the x-axis starts from the position xi = ......
A car moving along the x-axis starts from the position x_{i} = 2 m when t_{i} = 0 and stops at x_{f} = −3 m when t_{f} = 2 s. (a) Find the displacement, the average velocity, and the average speed during this interval of time. (b) If the car goes backward and takes 3 s to reach the starting point, then repeat part (a) for the whole time interval.
(a) The car’s displacement, see Fig. 3.3, is given by:
\Delta x=x_{\mathrm{f}}-x_{\mathrm{i}}=-3\ \mathrm{m}-2\ \mathrm{m}=-5\ \mathrm{m}The average velocity is then given by:
{\overline{{v}}}={\frac{\Delta x}{\Delta t}}={\frac{x_{\mathrm{f}}-x_{\mathrm{i}}}{t_{\mathrm{f}}-t_{\mathrm{i}}}}={\frac{-3\;\mathrm{m}-2\;\mathrm{m}}{2\;s-0\;s}}={\frac{-5\;\mathrm{m}}{2\,s}}=-2.5\;\mathrm{m/s}Since Δx and \bar{v} are negative for this time interval, then the car has moved to the left, toward decreasing values of x, see Fig. 3.3. The total covered distance is d = 5 m and the average speed is thus:
{\bar{s}}={\frac{{\mathrm{total \ distance}}}{{\Delta t}}}={\frac{d}{t_{\mathrm{f}}-t_{\mathrm{i}}}}={\frac{5\,{\mathrm{m}}}{2\,{\mathrm{s}}-0\,{\mathrm{s}}}}={\frac{5\,{\mathrm{m}}}{2\,{\mathrm{s}}}}=2.5\,{\mathrm{m/s}}In this case, \bar{s} is the same as \bar{v} (except for a minus sign).
(b) After the backward movement, the final position and final time of the car are x_{f} = 2 m and t_{f} = 2 s + 3 s = 5 s, respectively, while the total distance covered by the car is d = 5 m + 5 m = 10 m. As we know, the displacement involves only the initial and final positions and will be:
\Delta x=x_{\mathrm{f}}-x_{\mathrm{i}}=2\ \mathrm{m}-2\ \mathrm{m}=0Then, the average velocity will be:
{\overline{{v}}}={\frac{\Delta x}{\Delta t}}={\frac{x_{\mathrm{f}}-x_{\mathrm{i}}}{t_{\mathrm{f}}-t_{\mathrm{i}}}}={\frac{0}{5\mathrm{~s-0\mathrm{~s}}}}=0Finally, the average speed for the whole movement of the car will be:
{\bar{s}}={\frac{{\mathrm{total~distance}}}{\Delta t}}={\frac{d}{t_{\mathrm{f}}-t_{\mathrm{i}}}}={\frac{10\;{\mathrm{m}}}{5\;{\mathrm{s}}-0\;{\mathrm{s}}}}={\frac{10\;{\mathrm{m}}}{5\;{\mathrm{s}}}}=2\;{\mathrm{m/s}}As you can see, the average velocity is zero, while the average speed is 2 m/s, since the latter depends only on the total covered distance d.
