Question

A car moving with a velocity of 10m/s can be stopped by the application of a constant force F in a distance of 20m. If the velocity of the car is 30m/s. It can be stopped by this force in

• A. $\frac{20}{3}m$
• B. 20 m
• C. 60 m
• D. 180 m
• E. 7 m

Answer 1

Answer: (D)

180 m

Answer 2

The correct option is (D): 180 m.
The final velocity is zero, and the acceleration is −ve.
$0=u^2-2\times{a}\times{s}$
$s=\frac{u^2}{2a}$
S ∝ u2
If a car moving at a velocity of 10 m/s can be stopped by applying a constant force F over a distance of 20 m, then tripling the velocity will increase the stopping distance by a factor of 32=9.
New stopping distance = 9×20=180 m.