Speed Trap A stretch of highway has a speed limit of 70 mph. However, you will o

Question

Speed Trap A stretch of highway has a speed limit of 70 mph. However, you will only be pulled over on this stretch if your speed exceeds 75 mph (33.3 m/s). However, the reason that most people get pulled over is because this stretch of highway is a giant downhill segment which drops an elevation of 950 m during a linear stretch of road of 2200 m. The coefficient of friction on this road is 0.21. The mass your car is 750 kg. Your car is designed so that the air drag force on it in the horizontal direction is 870 N backwards and the air exerts a downward force on the car of 1720 N down. What is the kinetic energy your car when it is traveling at 33.3 m/s forward? How much work will friction do to the car during the 2200 m stretch of road? If you want to avoid speeding at the bottom of the massive hill then what velocity do you need to be traveling at the top of the hill (calculate it in m/s then convert t o mph)? Suppose before you got to the top of the hill (and it is flat before the top o c hill) that you were going 75 mph (33.3 m/s). What distance before the top of the hill would you want to let off the gas so that you could reach the velocity you found in C by the time your reach the hill without having to use your brake?

Explanation / Answer

A) velocity V = 33.3 m/s

KE = mv2/2 = 750*(33.3)2/2 = 415833.75 J

B) frictional force F = 0.21*mg = 0.21*750*9.8 = 1543.5 N

       distance = 2200 m

work done = 1543.5*2200 = 3395700 J

C) drop in elevation = 950 m

PE gained = mgh = 950*9.8*750 = 6982500 J

                   Work done against friction = -3395700 J

   Air drag    force      (Workdone)          = 870*2200 = -1914000 J

   Downward Air force (Wokdone)         = 1720* 950 = - 1634000 J

Net gain in energy = 6982500 -3395700 -1914000 - 1634000 = 38800 J

gain in vel = sqrt (2*38800/750) = 10.17 m/s

To avoid speed up the velocity at the top of the hill must be 33.3-10.17 = 23.13 m/s

                                                                               = 51.74 mph

D) vel = 75 mph = 33.53 m/s

KE = 750*33.532/2 = 421598 J

desired vel = 23.13 m/s

desired KE= 200624 J

loss of KE = 421598 -200624 = 220974 J.

This energy must be consumed by frictional force and he air drag on the vehicle.

Frictional force (workdone)= d*750*9.8*0.21

air drag   (workdone)         = d*870

Total work done = 2413.5 d = 220974

d = 91.56 m , the gas can be stopped so that the car will reach the desired speed before goingdown hill and do not over speed.

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