Question

A car moving with a speed of 50 km $h^{-1}$ can be stopped by brakes after atleast 6 m. If the same car is moving at a speed of 100 km $h^{-1}$ the minimum stopping distance is :

• A. 6 m
• B. 12 m
• C. 18 m
• D. 24 m

Answer 1

Answer: (D)

24 m

Given, $u=50km/hr=50\times \frac{5}{18}=\frac{250}{18}m/s$

$v=0s=6m\text{(at least)}$

By third equation by motion,

$v^2=u^2+2as0=(\frac{250}{18}\times \frac{250}{18})+2(-a)\left(s\right)0=192.90-12aa=\frac{192.90}{12}=16.07$

Now applying third equation of motion,

$v^2=u^2+2as⟹0=(100\times {\frac{5}{18})}^2+2(-16.07\left)\right(s)⟹0=771.60-32.14s⟹s=\frac{771.60}{32.14}=24.00$