# The position of a car as a function of time is given by x=(45m)+(−4.0m/s)t+(−8m/s^2)t^2.?

1- What is the acceleration of the car?

2-What distance does the car travel during the first 1.0 s?

3- What is the average velocity of the car between t=1.0s and t=2.0s?

### 1 Answer

Relevance

- billrussell42Lv 72 years ago
x = 45 − 4t − 8t²

x' = –4 – 16t

x'' = –16

1. –16 m/s²

2. at t=0, x = 45 – 0 – 0 = 45 m

at t=1, x = 45 – 4 – 8 = 33 m

Δx = 45–33 = 12 m

3. at t=2, x = 45 – 4•2 – 8•4 = 45 – 8 – 32 = 5 m

average velocity = Δx/Δt = (33–5) / 1 = 28 m/s

Still have questions? Get your answers by asking now.