Question

A car moving with a speed of $40km/hr$ can be stopped by applying breaks for $2$ km. If the same car is moving with a speed of $80km/hr$, what is the minimum stopping distance? Assume constant retardation.

• A. $8Km$
• B. $2Km$
• C. $4Km$
• D. $6Km$

Answer 1

The correct option is A $8Km$

$v^2-u^2=2as$

$\text{Substituting}v=0,u=40km/hr,s=2km$

$0-{40}^2=2\times a\times 2$

$a=-400km/h^2$

$\text{For}u=80km/h$,

$0-{80}^2=2\times (-400)\times s$

$s=\frac{6400}{800}=8km$