A car is traveling with a constant speed of 30.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with a constant acceleration. The car comes to a stop in a distance of 120 m. What was the acceleration of the car as it slowed down?

a = - 3.75 m/s²

negative sign indicates deceleration here.

Explanation:

In order to find the constant deceleration of the car, as it stops, we will use the 3rd equation of motion. The 3rd equation of motion is as follows:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration of the car = ?

Vf = Final Velocity = 0 m/s (Since, the car finally stops)

Vi = Initial Velocity = 30 m/s

s = distance covered by the car = 120 m

Therefore,

a = [(0 m/s)² - (30 m/s)²]/(2)(120 m)

a = - 3.75 m/s²

negative sign indicates deceleration here.

1/0.14nm = 10^10/1.4 a lot

be subject to a force andchange its motion