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

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  • 2 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

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