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
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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
Source: Myschool Past Questions Archive (Record #28557)