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

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

Answer Details

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