Answer to Question #88211 in Mechanics | Relativity for Amoo Emmanuel

Question

A train at 72km/h undergoes a uniform retardation of 2m/s when brakes are applied. Find the distance traveled from the place where the brakes were applied.

Answer 1

Let's first convert km/h to m/s:

$$v_0 = 72 \dfrac{km}{h} \cdot \dfrac{1000 m}{1 km} \cdot \dfrac{1 h}{3600 s} = 20 \dfrac{m}{s}.$$

We can find the distance traveled by the train from the place where the brakes were applied from the kinematic equation:

$$v_f^2 = v_0^2 + 2ad,$$

here, $v_0 = 20 m/s$ is the initial velocity of the train, $v_f = 0$ is the final velocity of the train, $a = -2 m/s^2$ is the deceleration of the train, $d$ is the distance traveled by the train from the place where the brakes were applied.

Then, we get:

$$d = \dfrac{v_f^2 - v_0^2}{2a},$$ $$d = \dfrac{(0 \dfrac{m}{s})^2 - (20 \dfrac{m}{s})^2}{2 \cdot (-2 \dfrac{m}{s^2})} = 100 m.$$

Answer:

$d = 100 m.$