Consider a vehicle with the velocity-time graph on the left.

a) FInd the velocity at 7s
40m/s

b) Find its acceleration during the first 5 seconds.
Acceleration =change in velocity/time taken
acc =( final velocity-initial velocity)/time
acc = (20m/s -0m/s)/5s = 4m/s/s

c) How far did the vehicle travel during the first 7 seconds.
This is given by the area under the graph.for the time period between t =0 and t=7.
Distance = velocity X time.

120m

d) When did the vehicle start its journey back?
The vehicle starts its journey back when the velocity becomes negative. In this case the journey back starts 8 seconds into the journey.

The image on the left shows the velocity time graph of a rocket that is launched vertically skywards on planet with no atmosphere.

a) What is the peak altitude reached by the rocket?
The peak altitude is the distance traveled by the rocket on its way up. This is given by the area under the graph between 0 and 7 seconds. Distance = velocity X time, in this case the altitude is 200m

b) What is the acceleration of the rocket during the 5 -7 second time period?
Acceleration is the change in velocity over time.
The change in velocity between 5 and 7 seconds is given by the expression
acc = (Final velocity -initial velocity) / 2
acc = (0m/s - 40m/s)/2s = - 20m/s/s
The negative sign indicates deceleration

c) What is the altitude of the rocket 5 seconds into the flight?
This is given by the area under the graph between 0 and 5 seconds. In this case it is
160m.

d) At what time did the engines of the rocket cease stop producing thrust?
Between 0 and 2 seconds into the flight the engines are producing enough thrust to overcome the force of gravity and lift the rocket skywards. Between 2 and 5 seconds the force produced by the engines just balances the force of gravity and allows the rocket to travel at constant speed. At the 5 second mark the engines have stopped and the force of gravity slows the rocket down.The rocket still travels skyward but is now slowing down.