What is the total distance the car covers?

Question

A car accelerates from rest for 10 s at a rate of 4.3 m/s². After driving at a constant speed for another 10 seconds, it starts to slow down at a rate of -3.4 m/s² until it comes to a stop. What is the total distance the car covers?

Answer 1

There are three parts to this.

1. Acceleration
2. Constant velocity
3. Negative acceleration

Δx = (1/2) a t² + v₀ t

Substituting in a = 4.3 m/s² at t = 10 s gives 215 m (starting from rest)

After accelerating from rest at 4.3 m/s² for 10s, the velocity is 43 m/s (Δv = a t). For 10s at 43 m/s the distance traveled is 430 m (from Δx = v t)

Δv = a t with Δv = -43 m/s and a = -3.4 m/s² gives t = 12.647 s stopping time.

Δx = (1/2) a t² + v₀ t with a = -3.4 m/s², t = 12.647 stopping time, and v₀ = 43 m/s because the car was already moving for part 3 gives 271.91 m.

Add the three for each part and get 916.91 m total distance.