Question

A car traveling with velocity v is decelerated by a constant acceleration of magnitude a. It travels a distance d before coming to rest. If its initial velocity were doubled, the distance required to stop would A) double as well. B) decrease by a factor of two. C) stay the same. D) quadruple.

Answer 1

Answer:

D) quadruple.

Explanation:

Assuming the same constant acceleration a in both cases, as we have as givens  the acceleration a, the distance d, and the initial velocity v, we can use the following kinematic equations in order to compare the distances:

vf² - v₀² = 2*a*d

As the final state of the car is at rest, the final velocity vf, is 0.

⇒ - v₀² = 2*(-a)*d ⇒ d =v₀² / 2*a

1) initial velocity v₀

d₁ = v₀² / 2 a

2 ) initial velocity 2*v₀

⇒ d₂ = (2*v₀)² / 2*a = 4*v₀² / 2*a ⇒ d₂ = 4* (v₀² / 2*a)

⇒ d₂ = 4* d₁

As the equation shows, the distance required to stop, if the initial velocity were doubled, the distance required to stop would quadruple.