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
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
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
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