Question

A car moves at a speed of $40$ km/h, it is stopped by applying brakes which produces a uniform acceleration of $-0.6$ $m/s^2$. How much distance will the vehicle move before coming to stop?

Answer 1

$40\frac{Km}{hr}=40\times 1000/(60\times 60)\frac{m}{s}(=11.1111\dots ).$

$a=\frac{“delta”v}{“delta”t}=\frac{(v_f-v_0)}{t}=\frac{(0-v_0)}{t},$

$t=\frac{-v_0}{a}=\frac{-11.11}{-0.6}=18.518\dots sec.,$

$d=v_0\times \frac{t}{2}=\frac{11.1111\times 18.518}{2}=102.9m$