i) V = u+ at Deceleration = u – v
0 = 20 + 2a OR t
a = - 10ms^-2 = (20 – 0)/2
=10ms^-2
ii)Stopping time = 2.2s
Before stopping = 0.2 x 20 = 4m
(10 – 202)/(2(-10))=400/20=20
20+4 =24m
Total time stop = 2.2 sec
S = ut +1/2 at^2
=(20 x 2.2) + 1/2 + 10 x 2.2^2
= 19.8m
Githiari answered the question on September 19, 2017 at 11:46
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