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A car moving at 30m/s slows down uniformly to a speed of 10m/s in a time of 5.0s. Determine a) the acceleration of the car, and b) the distance out moved in the third second |
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Answer» Answer: DATA: v1 = 30m/s v2 = 10 m/s a = ? t = 5 s d = ? SOLUTION: Solving for the acceleration of the car a = (10 m/s-30 m/s) / 5s a = -20 m/s / 5s a = -4 m/s^2 Solving for the DISTANCE traveled after the THIRD second d = v1 * t + 1/2at^2 d = 30 m/s * 3 s + -2m/s^2 * (3s)^2 d = 90 m + - 18 m d = 72 m Solving for the distance traveled after 2 seconds d = v1 * t + 1/2 at^2 d = 10 m/s * 2 s + -2 m/s^2 * (2 s)^2 d = 20 m + - 8 m d = 12 m Solving for the distance traveled in the third second d = distance moved after 3 s – distance moved after 2 s d = 72 m – 12 m d = 60 m
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