Question 1Draw a velocity-time graph. The area underneath must add up to 2000, and it will be some kind of triangle, with two of the vertices at
and
, where
is measured in seconds. The other vertex will lie somewhere in from
and will be
, as the train first accelerates. Let's call this vertex:
From the graph, you should be able to see that there is a maximum velocity attained at that last vertex I described.
The velocity-time graph is made up of two linear sections - an accelerating part, and a decelerating part. The acceleration is represented by the gradients of these linear sections. Since the deceleration (retardation) is double the acceleration (
vs
), then therefore the time taken for the first section (flatter gradient) is twice as much as the second section (steeper gradient).
So, therefore,
(two-thirds of the way), since the first part of movement takes 2 times longer than the second part (i.e.: a 2:1 ratio -
).
Also, we know the triangles have a total area of
Therefore, using
and
, then: