velocity-time graph of journey

a) `text(distance)= "area of trapezium"`

`=((a+b)h)/2`

`=((80+30)(40))/2`

`=2200\ text(m)`

b) `text(average speed)=text(distance travelled)/text(time taken)`

`=2200/80`

`=27.5\ text(ms)^-1`

c) We need to find the time when the area of the trapezium is half of its original area, or `1100 text(m)^2`, as shown in the graph.

velocity-time graph of journey 2

The base of this unknown trapezium has length `t`, and the top of the trapezium will have length `t-20`. So we have:

`"area of trapezium" =((a+b)h)/2`

`1100=((t+[t-20])40)/2=20(2t-20)`

`55=2t-20`

`75=2t`

`t=37.5 text(s)`

So it will take `37.5\ "s"` to cover half the distance.