5. A car moving at +12.0 m/s at the bottom of a hill, coasts up the hill with a uniform acceleration rate of -1.60 m/s'.

a. What is its displacement after 6.00 seconds?
b. What is its displacement after 9.00 seconds?

1 answer

To solve these problems, we can use the equations of motion:

1. Displacement equation:
Δx = v₀t + 0.5at²
where Δx is displacement, v₀ is initial velocity, t is time, and a is acceleration.

Given:
Initial velocity, v₀ = +12.0 m/s
Acceleration, a = -1.60 m/s²

a. Displacement after 6.00 seconds:
t = 6.00 s

Δx = (12.0 m/s)(6.00 s) + 0.5(-1.60 m/s²)(6.00 s)²
= 72.0 m - 0.5(1.60 m/s²)(36.0 s²)
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