1. ## speed

According to Car and Driver, an Alfa Romeo going 70 mph requires 177 feet to stop. Assuming that the stopping distance is proportional to the square of the velocity, find the stopping distance required by an Alfa Romeo going at 45 mph and at 135 mph.
At 45 mph, stopping distance equals?
At 135 mph, stopping distance equals?

2. Originally Posted by cwarzecha
According to Car and Driver, an Alfa Romeo going 70 mph requires 177 feet to stop. Assuming that the stopping distance is proportional to the square of the velocity, find the stopping distance required by an Alfa Romeo going at 45 mph and at 135 mph.
At 45 mph, stopping distance equals?
At 135 mph, stopping distance equals?
If stopping distance is proportional to the square of the velocity then we know that $d = kv^2$ where k is some constant. So find k from the 177 ft and 70 mph data, then use that constant to calculate the distance for the other two speeds.

For example,
$k = \frac{177}{70^2} \approx 0.036122$
(Check the units here. We are requiring that the distance is in feet and the speed is in mph in order to use this constant!)

So the stopping distance for 45 mph is
$d = kv^2 = (0.036122)(45^2) \approx 73.15~ft$

-Dan