You are a police officer. Your squad car is at rest on the shoulder of an inters

Question

You are a police officer. Your squad car is at rest on the shoulder of an interstate highway when you notice a car fitting the description given in an all-points bulletin passing you at its top speed of 85 mi/h. You jump in your car, start the engine, and find a break in the traffic, a process which takes 25 s. You know from the squad car's manual that when it starts from rest with its accelerator pressed to the floor, the magnitude of its acceleration is a = a0 - bt^2 , where a0 = 2.5 m / s^2 and b = 0.0028 m / s^4 , until a0 = bt^2 and then remains zero thereafter. Can you catch the car before it reaches the next exit 5.3 mi away?

Explanation / Answer

Let the police officer catch the car after time t

85 mi/h = 85*0.44704 = 37.9984m/s                                (1 mi/h = 0.44704 m/s)

distance travelled by the car in t sec = speed* time = 37.9984t

magnitude of acceleration of the police officer = a0 - bt^2

                                                  = 2.5 -0.0028t^2

the police car accelrates for t sec   t = sqrt(2.5/0.0028) = 29.88 sec

velocity of the police car = integral ( 2.5 - 0.0028t^2) from t = 0 to 29.88

v = 2.5t - 0.0028t^3 / 3 from t = 0 to 29.88

v = 49.8 m/s

distnace travelled by the car in t sec = 49.8(t-25)

49.8(t-25) = 37.9984t

t = 105.48

so distance after which the car is caught = 37.9984*105.48 = 4008.20 m = 4km

5.3 mi = 5.3*1.60934 = 8.53km

so , the car will be caught before the next exit 5.3 mi away

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