Equation relating to stopping distance to velocity

Question

Driving on the Autobahn in Germany, you have to brake suddenly. Let’s assume that the stopping distance of a car varies directly as the square of the velocity of car when the brakes are applied. A car moving at 85 mph can stop in 367 feet. 

a. Write an equation that relates the stopping distance to the velocity. 

b. What is the stopping distance of your car if you were moving at 95 mph?

Answer 1

First we need to set up the relation between distance and velocity.

Let's set:

s=distance

v=velocity

k=constant of variation

Since distance is directly related to the square of the velocity our equation should be: (Remember that when something is direct variation we multiply by k)

s=kv²

Using the given distance and velocity from the initial word problem we can use our equation to solve for k.

367 = k(85²)

Solving for k: k=0.0508

a.) our equation for this situation should be given as: s = 0.0508v².

b.) plug 95 mph into our equation with the known k and solve for s. s = 0.0508(95²) = 458.5 ft.

Hope this explanation helps.

Answer 2

a. Assuming the stopping distance is y, and initial velocity is x. The stopping distance can be expressed as

y = a x²

where a coefficient of the equation to be decided.

b. The stopping distance for 85 mph is 367 feet, so

367 = a 85²

and the stopping distance for 95 mph can be expressed similarily as:

d = a 95²

thus

d/367 = 95²/85²

d = 367 (95/85)² = 458 ft