Hi,
At the maximum speed, a = 0, so you are right that T =
4.
Next, we need to find v by integration, i.e. v = 8t - t^2
+ c.
When t = 0, v = -7, so v = 8t - t^2 - 7.
Sketch a graph of v against t. This is an inverted
u-shaped curve.
We only consider the region of the graph between t = 0
and t = 4.
During this period of time, distance travelled
= - integral { 0, 1 } v dt + integral { 1, 4 } v dt ---
(1)
Beyond t = 4, we need to use distance travelled = ut +
1/2 at^2, because we are told that the previous acceleration
formula applies only for 0 < t < T.
From the maximum velocity of 9 m/s, it will take 6s to
drop to 0 m/s at the acceleration of -1.5 m/s^2.
Now, distance travelled = (9)(6) + 1/2 (-1.5)(6^2) = 27
--- (2)
Hence total distance = (1) + (2).
Thanks!
Cheers,
Wen Shih
