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

95m

100m

500m

905m

1000m

Correct answer is A

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

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