1)A car slows down from 22 m/s torest in a distance of 91 m. What wasits acceler

Question

1)A car slows down from 22 m/s torest in a distance of 91 m. What wasits acceleration, assumed constant?
1 m/s2
2)A stone is thrown vertically upward with a speed of14.0 m/s from the edge of a cliff55.0 m high

(a) How much later does it reach the bottom of thecliff?
1 s
(b) What is its speed just before hitting?
2 m/s
(c) What total distance did it travel?
3 m

1)A car slows down from 22 m/s torest in a distance of 91 m. What wasits acceleration, assumed constant?
1 m/s2
2)A stone is thrown vertically upward with a speed of14.0 m/s from the edge of a cliff55.0 m high

(a) How much later does it reach the bottom of thecliff?
1 s
(b) What is its speed just before hitting?
2 m/s
(c) What total distance did it travel?
3 m

Explanation / Answer

1. v^2=u^2+2as 0 = 22^2 + 2a(91) a = - 22^2/(2*91) = -2.66 m/s^2 2. u=14, a= -g = -9.8, a) x = ut + (1/2)at^2 -55 = 14t - (.5*9.8)t^2 -55 -14t + (.5*9.8)t^2 = 0 2 solutions of this quadratic equation are t = -2.2 and 5.1 -3.2 is unphysical, so the answer is 5.1 sec. b) v = u + at = 14 - 9.8*5.1 = -36 m/s c) max height H above the cliff is given by v^2 = 0 = u^2 + 2aH = 14^2 - 2*9.8*H H = 10 m total distance = 10 + 10 + 55 = 75 m

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