# A 2230 kg car traveling to the west at 17.5 m/s slows down uniformly. How long would it take the car to come to as top if the force on the car is 8810 N to the east?

### 1 Answer | Add Yours

The initial velocity of the car is 17.5 m/s towards the West. The mass of the car is 2230 kg. A force of 8810 N acting towards the East acts on the car. The acceleration on the car due to the force applied is 8810/2230 m/s^2 towards the East. This decelerates the car and its velocity decreases.

For an object traveling at an initial velocity U and on which there is a constant acceleration of A, the final velocity after t seconds is V = U + A*t

Here U = 17.5 m/s and A is 3.95 m/s^2 towards the East or -3.95 m/s^2 towards the West. The final velocity is V = 0.

Substituting these values in V = U + A*t

=> t = (V - U)/A

=> t = (0 - 17.5)/(-3.95)

=> t = 4.429 s

**The time taken by the car to come to a halt is 4.429 s**