Answer to Question #134059 in Mechanics | Relativity for David P. Costello

Question

An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first sees the car, the locomotive is 200 m from the crossing and its speed is 20 m/s. If the engineer’s reaction time is 0.49 s, what should be the magnitude of the minimum deceleration to avoid an accident? Answer in units of m/s 2 .

Answer 1

Explanations & Calculations

• The train travels some distance during the engineer's reaction time & if that is S₁ then,

$\qquad\qquad \begin{aligned} \small S_1 &= \small ut\\ &= \small 20ms^{-1}\times 0.49s\\ &= \small 9.8m \end{aligned}$

• Then a distance of $\small (200m-9.8m =) 190.2m$ is left for the engineer to stop the train under any acceleration which the train is capable of.
• Assuming that this deceleration remains constant throughout & applying $\small V^2 =U^2+2as$ , deceleration could be found.

$\qquad\qquad \begin{aligned} \small s& = \small \frac{V^2-U^2}{2a}\\ \small s &\leq \small 190.2 \,\,\,:\text{to avoid the impact}\\ \small \frac{0^2-20^2}{2\times a}& \leq \small 190.2 \\ \small a & \geq \small -1.052ms^{-2}\,\,\, : \text{negativity = deceleration}\\ \small |a|_{min} & =\small \bold{1.052ms^{-2}} \end{aligned}$

• Any deceleration of greater value than this stops the train at some distance (of greater safety) before the impact point.