raking. akes a Ql I call to report that bc?s lost his brakes and his atrol car p

Question

raking. akes a Ql I call to report that bc?s lost his brakes and his atrol car picks up the call and races to get ahead of the he slows up and matches his speed to the Ferrari. As jumpers with him, the policeman brakes as hard as he this time, they arc traveling at 60 mph dosn a 15 grade. t is 1.940 lb and the police cruiser?s curb weight is ser can normally brake from 60 mph in 250 ft On lcvc the policeman to bring the two cars to rest? traveled during this time? cruiscr pulls in front of him and does tI

Explanation / Answer

A car weighing 1940lb has lost its brakes and is unable to come to a stop. A police cruiser of 3900lb gets ahead of it so that the two cars are now bumper to bumper. The combined cars are now traveling at 60mph down a 15 degree grade slope. If the police cruiser can normally brake from 60mph to rest in 250ft if it is on level ground: A) How long will it take for the police cruiser to bring both cars to a stop? B) How far will they have traveled during this time? Ignore friction. 2. Relevant equations F= ma V^2 - Vo^2 = 2a?x 3. The attempt at a solution The first thing i did was convert everything into SI units for simplicity. Speed of Cars 26.8 m/s Mass of car 881kg Mass of police cruiser 1770kg Weight of car 8643N Weight of police cruiser 17360N Weight of Combined Cars 26000N I then determined the breaking force Fb of the police cruiser's brakes by determining first the acceleration of caused by the brakes, and then finding the force using F=ma [V^2 - Vo^2] / [2?x] = a ab= [0 - (26.8m/s)^2]/[2(76.3m)] = -4.71m/s^2 Fb=mab F=(1770kg)(-4.71m/s^2) = -8337N So the force applied by the brakes of the police is -8337N.

cars going down a 15 degree slope. So if the cars are currently moving at a speed of 26.8m/s down the slope, this means they are not accelerating ?Fx = Wx - Fbx

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