On a dry asphalt road, a car's stopping distance varies directly as square of its speed

Question

On a dry asphalt road, a car's stopping distance varies directly as square of its speed. A car traveling at 45 miles per hour can stop in 135/2 feet. What is the stopping distance for a car traveling at 60 miles per hour?

Answer 1

Osiris,

 

The stopping distance is 240.2 ft at 60 mph if it is 135.2 ft at 45 mph.  It's half of that (120.1 ft) at 60 mph if 135/2 = 67.5 ft is used at 45 mph. Road & Track magazine shows 60-0 mph stopping distances of around 120 ft for most good cars it tests.  Hence, the preferred interpretation of the input is 135/2 = 67.5 ft.

 

Here's the solution.

 

Let   s = the car's speed when the brakes are applied (in feet/sec).

       d = stopping distance (in feet).

       c = the conversion constant of speed from the given mph to ft/sec = (5200 ft/mi) /(3600 sec/hr) = 1.444

       k = the constant of proportionality of d to s-squared.

Then,

At 45 mph, s = 1.444(45) = 65 ft/sec

At 60 mph, s = 1.444(60) = 86.64 ft/sec.

 

And    d / s²  = k according to the first sentence in the problem statement.

 

So at 45 mph,  k = 67.5 ft/ (65 ft/sec)² = 67.5 / 4225 = 0.016

 

Then at 60 mph, d = ks2 = 0.01598 (86.64)2 = 0.016(7506.49) = 120.1  ft.

 

Answer 2

Osiris,

 

This type of problem is very common in physics and physical science course and comes from the fact that we can equate the amount of work done by friction is equal to the change in kinetic energy (motion) of the car.  The work done by friction is equal to the frictional force x the distance required for the car to come to a stop.

 

Simply:  F x d = ½mV²

 

m is the mass of the car, and F is the frictional force which we can assume to be constant.

 

If we write this equation twice, once for the 1st speed and distance, and again for the second speed and distance we will have:

 

½mV₁² = F·d₁ and ½mV₂² ₌ F·d₂ 

 

Since F and ½m are constant, if we divide equation 2 by equation 1 they will cancel out and we have:

 

(V₂/V₁)² = d₂/d₁

 

multiply both sides by d₁ and get:

 

d₂ = (V₂/V₁)²·d₁

 

 

(V2/V1)² = (60/45)² = (4/3)²

 

so we have:

 

d₂ = (4/3)²·(135'/2) = (16/9)·(135'/2) = (8)·(15') = 120'