Question

A 1,500 − ?? car is traveling at 100 ??/h. The driver sees a slow truck ahead and slows down to 60 ??/h. a. Find the initial and final energy of the car and the work done on it to slow it down. b. If the car completes its deceleration in 9.0 ?, find the mean braking power. c. A typical car can be made to stop from 100 ??/h in a minimum distance of 60 ? if the pavement is dry. Assuming that this specification applies to this car, in part b), find the maximum braking power of the car and compare this value with the power calculated in b) (how much bigger/smaller is it).

Answer 1

a)

initial kE

kEi = 0.5* 1500* (100* 5/18)^2 = 5.79* 10^5 J

final kE

kEf = 0.5* 1500* ( 60* 5/ 18)^2 = 2.083* 10^5 J

workdone

W = change in kE

W = (5.79 - 2.083)* 10^5 = 3.707* 10^5 J

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b)

P = W/t = 3.707* 10^5 / 9 = 41185 W

c)

acceleration of the car

a = v^2 / ( 2d) = (100* 5 / 18)^2 / ( 2* 60) = 6.43 m/s^2

time it takes to stop

t = v/a = ( 100* 5 / 18) / 6.43 = 4.32 s

power

Po = 0.5* 1500* (100*5/18)^2 / 4.32

Po = 1.34* 10^5

applied power = 41185 /(1.34* 10^5) * peak power

P = 0.307 * Po

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