A 2230 kg car traveling to the west at 17.5 m/s slows down uniformly.The force on the car is 8810 N to the east.What is the car’s displacement during the time it takes to stop?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The distance D traveled by an object with an initial velocity U m/s and which is accelerated at A m/s^2 till the final velocity becomes V m/s, is given by the relation V^2 - U^2 = 2*A*D

Here, the initial velocity of the car is 17.5 m/s towards the west. A force of 8810 N acts on it towards the East. As the mass of the car is 2230 kg, the acceleration acting on it is -8810/2230 = -3.95 m/s^2 towards the west. The velocity of the car is decreased due to this and it stops after traveling a distance D.

Substituting the values obtained

0^2 - 17.5^2 = 2*(-3.95)*D

=> D = 17.5^2/(2*3.95)

=> D = 38.75 m

The displacement of the car during the time it takes to stop is 38.75 m.