If a car starts from rest and moves with a uniform acceleration of 10ms-2 for ten seconds, the distance it covers in the last one second of its motion is
A.
95m
B.
100m
C.
500m
D.
905m
E.
1000m
Correct Answer: Option A
Explanation
Distance in 105, using, x = ut + \(\frac{1}{2}\)at2
x = (0 x 10) + (\(\frac{1}{2}\) x 10 x 10 x 10) = 500m
distance in 9s; x - (0 x 9) + (\(\frac{1}{2}\) x 10 x 9 x 9) = 405m
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