Question

A 1,500-kilogram car travelling at 10 meters per second (about 22 mph) strikes a parked car on the side of the road. If the car comes to a stop in half of a second, what force is exerted on the parked car by the moving car?

Answer 1

The force exerted on the parked car by the moving car is –30000 N

To solve this question, we'll begin by calculating the deceleration of the moving car. This can be obtained as follow:

Initial velocity (u) = 10 m/s

Final velocity (v) = 0 m/s

Time (t) = 0.5 s

Deceleration (a) =?

$a = \frac{v - u}{t} \\\\ a = \frac{0 - 10}{0.5}\\\\a = \frac{- 10}{0.5}$

a = - 20 m/s

Finally, we shall determine the force exerted by the moving car on the parked car. This can be obtained as follow:

Mass of moving car (m) = 1500 Kg

Acceleration (a) = - 20 m/s

Force (F) =.?

F = ma

F = 1500 × (-20)

F = -30000 N

NOTE: The negative sign indicate force is in opposite direction to the parked car.

Therefore, the force exerted on the parked car by the moving car is –30000 N

Learn more: brainly.com/question/15430805