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
distance in last is = difference
= 500 – 405
= 95m