You are here

$d_2=2.25d_1$

Giancoli 6th Edition, Chapter 6, Problem 21

(2:34)

Chapter 6, Problem 21 is solved.

View sample solution

Transcript for this Giancoli solution

We’re going to see how the breaking distance of a car changes with change in speed. So we’ll start by saying the change in kinetic energy of the car is the net force times the distance and the force being provided by the… well, the brakes, but really the friction on the road. So the change in kinetic energy is one half ‘mv’ final squared minus one half ‘mv’ initial squared. The final velocity being zero ‘cause the car comes to a stop. So we can say that ‘F’ net times distance is negative a half ‘mv’ initial squared. And then we can solve for ‘d’. So we’ll divide both sides by ‘F’ net; divide by ‘F’ net and we get that ‘d’ equals negative ‘mv’ initial squared divided by two ‘F’ net. Okay. So on the one hand we have distance one when it has some particular speed, ‘v’ one, ‘v’ initial one squared divided by two ‘F’ net. And on the other hand we have the breaking distance when the speed is increased by 1.5 or 50% more. So we have one… negative ‘m’ times ‘v’ initial one times 1.5. Multiplying by 1.5 is the same as adding 50%. And we square that divided by two ‘F’ net. This is the same as 2.25. That’s one and a half squared is, 2.25 times negative ‘mv’ initial one squared divided by two ‘F’ net. And this… these factors in brackets, that’s what ‘d’ one is. That was the distance in the first case. So we can see that ‘d’ two equals 2.25 ‘d’ one. So with an increase in speed of 50% you’re more than doubling the stopping distance. So you get a sense from this equation of how much speed is a factor in car safety.