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A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 27.0 m/s by a 7850 N braking force acting opposite the car’s motion.
What is the car's velocity after 2.52s?
How far does the car move during the 2.52 s?
How long does it take the car to come to a complete stop?

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MathPhys

Answer:

19.1 m/s

58.1 m

8.60 s

Explanation:

Take north to be positive and south to be negative.

Use Newton's second law to find the acceleration.

∑F = ma

-7850 N = (2500 kg) a

a = -3.14 m/s²

Given:

v₀ = 27.0 m/s

a = -3.14 m/s²

Find: v given t = 2.52 s

v = at + v₀

v = (-3.14 m/s²) (2.52 s) + 27.0 m/s

v = 19.1 m/s

Find: Δx given t = 2.52 s

Δx = v₀ t + ½ at²

Δx = (27.0 m/s) (2.52 s) + ½ (-3.14 m/s²) (2.52 s)²

Δx = 58.1 m

Find: t given v = 0 m/s

v = at + v₀

0 m/s = (-3.14 m/s²) t + 27.0 m/s

t = 8.60 s

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