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

Question 1 Report

95m

100m

500m

905m

1000m

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

Read lesson note on Motion (JAMB)

Motion

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