A 1200-kg car, initially at rest, is initially accelerated at 4.2m/s^2. After th

Question

A 1200-kg car, initially at rest, is initially accelerated at 4.2m/s^2. After the car moves 730m, the engine is cut off. The car then moves a distance of 1440 before coming to a stop Assume the road to be level and that g = 10 N/kg. a) What is in m/s, the magnitude of the velocity of the car at me end of the first 730m? b) What is in m/s^2, the magnitude of the acceleration of the car while it is slowing down c) What is in N, the magnitude of the friction force (the sum of all forces that are slowing the car down) on the car While it is slowing down? d) Assuming that the car was subject to the same frictional forces While it was speeding up, what Is the magnitude of the push the road exerted on the car during the first 730m on the motion?

Explanation / Answer

A. 730 = 1/2a *t2 => t = 18.64s   => v = at = 4.2 * 18.644 = 78.3m/s

B. v2 - u2 = 2as      u = 0, v= 78.3 , s =1440     => a = 6130.89 /2*1440 => a= 2.12 m/s2

C. F= ma => 1200* 2.12 = 2554N

D. Magnitude of push = Force which cancels Friction force + the acceleration force = 2554 + 4.2*1200 = 7594

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