(2) t = 2t2 – 8t + 5
Given, y(0) = +5m. From the graph,
Velocity = –8 ms-1 at t = 0
∴ v(0) = -8 ms-1
Now, v – t graph is a straight line. Therefore acceleration is a constant a = slope of v vs t graph.
= + 8/2 = +4 ms-2
The final position at t = t can be written as