Question

A car moves at 12 m/s for 30.0 seconds. It then accelerates at 1.5 m/s2 for 8.00 seconds.

Finally, it continues on at this top speed for another 12.0 seconds. Calculate the net
displacement during the whole time interval.

Answer 1

Over the first 30.0 s, the undergoes a displacement of

(12 m/s) * (30.0 s) = 360 m

Over the next 8.00 s, the car accelerates from 12 m/s to a top speed of

12 m/s + (1.5 m/s²) * (8.00 s) = 24 m/s

and over this time interval, it is displaced an additional

(12 m/s) * (8.00 s) + 1/2 (1.5 m/s²) * (8.00 s)² = 144 m

For the last 12.0 s, the car moves at a constant speed of 24 m/s to cover a distance of

(24 m/s) * (12.0 s) = 288 m

So the car's net displacement is 360 m + 144 m + 288 m = 792 m. (The net displacement is the same as distance in this case because the car moves in only one direction.)