A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is initially traveling at speed 7.0v? Assume that the acceleration due to the braking is the same in both cases.

Express your answer using two significant figures.

This is a kinematics problem, let's use the equation vf² = vi² + 2ad, where vf is final velocity, vi is initial velocity, a is acceleration and d is the distance we went. Considering the 1st scenario where we're going v initially, vi = v, vf = 0 (because it stops at the end), d = d₁ (because we'll go a different d when we go 7 times faster initially in the next part) and a can just be a since we don't have a value for it and it's always the same. Using this, let's solve for a (so we can eliminate it) and get a = - v²/2d₁ (acceleration should be negative if you're wondering about the sign, as we're slowing down). Setting up the same equation in the 2nd scenario, we use vi = 7v, vf = 0, d = d₂ and a is still a. Solving for a, we get a = - (7.0v) ²/2d₂. Now we can set our a expressions equal to each other, so - v²/2d₁ = - (7.0v) ²/2d₂, clean cancels on the 2, -, and v², solving for d₂ (our distance going 7 times our than initial velocity in the 1st scenario) we get d₂ = 49d₁ (don't forget to square the 7), meaning we go 49 times further when we goes 7 times faster initially.

Answer: 49 d

Explanation:

1) Stop distance ⇒ final velocity = 0

2) Work done by the brake system = change in kinetic energy: W = ΔKE

3) W = force * distance = F*d

4) ΔKE = [1/2] mv² ← taking in account final velocity = 0

5) ∴ F*d = [1/2]mv²

6) Since, acceleration and mass are constants, F is also constant.

7) When velocity is v₂ = 7.0v, you get d₂:

F*d₂ = [1/2]m (7.0) ²

8) Divide the equation from 7) by the equation from 5)

d₂ / d = (7.0v) ² / v² ⇒ d₂ = 49.0 d

Use two significant figures: d₂ = 49 d← answer