Question 8
An object starting from rest is subjected to a uniform acceleration for 4 s and reaches a velocity of 12 m s-1. It then moves for a further 4 s at that uniform velocity of 12 m s-1. Finally it is uniformly retarded and comes to rest in 2 s.
What is the object's total displacement over the 10 seconds?
The right answer is the one marked above.
84 m. Work out the three stages separately and add them.
First 4 s — average velocity × time = (0 + 12) ÷ 2 × 4 s = 6 × 4 = 24 m
Second 4 s — uniform velocity × time = 12 m s-1 × 4 s = 48 m
Last 2 s — average velocity × time = (12 + 0) ÷ 2 × 2 s = 6 × 2 = 12 m
total displacement = 24 m + 48 m + 12 m = 84 m
So the object finishes 84 m from where it started, in a straight line. Leaving out one of the stages is the commonest slip here.