Question

When a high-speed passenger train travelling at 161 km/hrounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D=676ahead (Fig.2-32). The locomotive is moving at 29 km/h. The engineer of the high speed train immediately applies the brakes. (a)What must be the magnitude of resulting constant deceleration if a collision is to be just avoided? (b)Assume that engineer is atx=0when, at t=0, he first spots the locomotive. Sketch x(t) curves for the locomotive and high speed train for the cases in which a collision is just avoided and is not quite avoided.

Short answer

The magnitude of the acceleration is 0.994 m/s²

Step-by-step solution

Given information

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$\mathrm{D}=676\mathrm{m}$

Understanding the concept of average velocity

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,

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

(a) Determination of deceleration of train

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

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

Hence,

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

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

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

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 $-0.994\mathrm{m}/{\mathrm{s}}^2$.

(b) Graph for the collision which is just avoided

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.