91影视

Velocity of high speed train${\mathrm{v}}_{\mathrm{t}}=161\mathrm{km}/\mathrm{hr}$

Velocity of locomotive${\mathrm{v}}_{\mathrm{I}}=29.0\mathrm{km}/\mathrm{hr}$

Distance between train and locomotive$\mathrm{D}=676\mathrm{m}$

The problem deals with the average velocity. It is the change in displacement divided by the time intervals in which the displacement occurs. Using the formula for the average velocity and acceleration, the magnitude of the acceleration can be found.

Formula:

The average velocity is given by,

$\frac{{\mathrm{v}}_{\mathrm{t}}+{\mathrm{v}}_{\mathrm{I}}}{2}=\frac{鈭�\mathrm{x}}{\mathrm{t}}$

Let $v_t$be the initial velocity of the train and $v_I$be the initial velocity of the locomotives.

Let $鈭�x$be the distance between them, and the forward distance traveled by the locomotives during time t is $v_{I}t$.

Hence,

$$\begin{gathered} \frac{v_t+v_I}{2}=\frac{鈭�x}{t} \\ \frac{v_t+v_i}{2}=\frac{D+v_{I}t}{t} \\ \frac{v_t+v_i}{2}=\frac{D}{t}+v_I \end{gathered}$$

As $t=\frac{v_{I鈥�}{v}_t}{a}$, above equation becomes as,

$$\begin{gathered} \frac{v_t+v_I}{2}=\frac{Da}{v_{I鈥�}{V}_t}+v_I \\ a=\left(\frac{v_t+v_I}{2}-v_I\right)\left(\frac{v_t+v_I}{D}\right) \\ a=-\frac{1}{2D}{\left(v_t+v_I\right)}^2 \\ a=-\frac{1}{2\left(0.676\mathrm{km}\right)}{(29\frac{\mathrm{km}}{\mathrm{h}}-161\frac{\mathrm{km}}{\mathrm{h}})}^{2}a=-12888\mathrm{km}/\mathrm{h} \end{gathered}$$

As $$\begin{gathered} 1\mathrm{km}=1000\mathrm{m} \\ 1\mathrm{h}=3600\mathrm{s} \end{gathered}$$

$$\begin{gathered} \mathrm{a}=\left(-12888\frac{\mathrm{km}}{{\mathrm{h}}^2}\right)\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right){\left(\frac{1\mathrm{h}}{3600\mathrm{s}}\right)}^2 \\ =-0.994\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, magnitude of the acceleration is $-0.994\mathrm{m}/{\mathrm{s}}^2$.

Distance is along y axis in meters and time t is along the x axis. The motion of the locomotives is shown by top lines and the motion of the passenger train is shown by bottom curve.