Question

An electric train starting from rest attains a maximum

speed of 200 kmph in 20 seconds. Determine :

(i) Its acceleration assuming it to be uniform

(ii) Distance covered during this time period

(iii) Its velocity 15 seconds after staring from rest

Answer 1

From the formula for univariate movement

$v= v_0+a \cdot t$

where by the condition of the problem $v= 200 {km/h}=55.56{m/s}$ $v_0=0$

1)define acceleration

$a=\frac{v-v_0}{t}=\frac{55.56-0}{20}=2.778[m/{s^2}]$

2) Distance traveled is

$s= \frac{a \cdot t^2}{2}=\frac{2.778 \cdot 20^2}{2}=555.6m$

3)Its speed after 15 seconds from the start of movement

$v=a \cdot t_1=2.778 \cdot 15=41.67{m/s}$