JAMB - Physics (1985 - No. 5)
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
Paaiškinimas
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
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