Question

A train slows from 108 km per hour with a uniform acceleration of 5 m per second square. How long will it take to reach 18 km per hour and the what is the distance covered.

Answer 1

By defintion, the acceleration is given as follows:

$$a = \dfrac{v_f - v_i}{t}$$

where $v_f = 18km/h = 5m/s, v_i = 108km/h = 30m/s$ are the final and initial speeds respectively, and $t$ is the time required to reach from $v_i$ to $v_f$. Solving for $t$ and substituting $a = -5m/s^2$ (minus since it decelerates), find:

$$t = \dfrac{v_f-v_i}{a} = \dfrac{30m/s - 5m/s}{-5m/s^2} =5s$$

The distance is given by the kinematic equation:

$$d = v_it + \dfrac{at^2}{2} = 30m/s\cdot 5s - \dfrac{5m/s^2\cdot (5s)^2}{2} = 87.5m$$

Answer. 5s and 87.5m.