Question 3.6: The remote-controlled truck shown in Fig. 3.12 moves along t......
The remote-controlled truck shown in Fig. 3.12 moves along the x-axis with a constant acceleration of −2 m/s². As it passes the origin, i.e. x_{o} = 0, its initial velocity is 14 m/s. (a) At what time t′ and position x′ does v′ = 0 (i.e. when the truck stops momentarily)? (b) At what times t_{1} and t_{2} is the truck at x = 24 m, and what is its velocity then?
(a) Given v_{o} = 14 m/s, v′ = 0, and a = −2 m/s², we can find t′ by using v′ = v_{o} + a t′ as follows:
t^{\prime}={\frac{v^{\prime}-v_{\mathrm{{o}}}}{a}}={\frac{0-14\;{\mathrm{m}}/s}{-2\;{\mathrm{m}}/s^{2}}}=7\;sTo find the position x′ we can use v^{\prime2}=v_{\circ}^{2}+2\ a\ (x^{\prime}-x_{\circ}), since we are given v_{o} = 14 m/s, v′ = 0, x_{o} = 0, and a = −2 m/s². Thus:
x^{\prime}=x_{o}+{\frac{v^{\prime \ 2}-v_{o}^{2}}{2\,a}}=0+{\frac{0-(14\,\mathrm{{m/s}})^{2}}{2\times(-2\,\mathrm{{m/s}}^{2})}}=49\,\mathrm{m}(b) Using x = 24 m, x_{o} = 0, v_{o} = 14 m/s, and a = − 2 m/s² in x − x_{o} = v_{o} t + \frac{1}{2} a t^{2}, we find, after omitting the units temporarily, that:
24-0=14\;t+{\frac{1}{2}}\;(-2)\;t^{2}\;\;\;\Rightarrow\;\;\;t^{2}-14\;t+24=0Solving this quadratic equation yields:
t={\frac{14\pm{\sqrt{(-14)^{2}-4\times1\times24}}}{2\times1}}={\frac{14\pm10}{2}}\ \ \Rightarrow\ \ t= \left\{\begin{matrix} t_{1}=2 \ s \\ t_{2}=12 \ s\end{matrix} \right.Thus, t_{1} = 2 s is the time the truck takes from the origin to the position x = 24 m.
Furthermore, t_{2} = 12 s is the time the truck takes from O, passing the point x = 24 m, reaching the point x′ = 49 m and returning back to x = 24 m.
For x = 24 m, v_{o} = 14 m/s, a = −2 m/s², and t_{1} = 2 s, we use the formula v = v_{o} + a t to get v_{1} as follows:
v_{1}=v_{o}+a\;t_{1}=14\;{\mathrm{m}}/{\mathrm{s}}+(-2\;{\mathrm{m}}/{\mathrm{s}}^{2})\times(2\;{\mathrm{s}})=10\;{\mathrm{m}}/{\mathrm{s}}Also, for x = 24 m, v_{o} = 14 m/s, a = −2 m/s², and t_{2} = 12 s, we use the formula v = v_{o} + a t to get v_{2} as follows:
v_{2}=v_{o}+a\ t_{2}=14\ m/s+(-2\ \mathrm{m/s^{2}})\times(12\ \mathrm{s})=-10\ \mathrm{m/s}Observe that the two speeds are equal, i.e. |v_{1}|=|v_{2}| = 10 m/s.
In this example, we do not pay any attention to the cause of this constant acceleration, but this will be clarified later on when we study the dynamical aspect of mechanics.