Math Is Fun Forum

Thimira Rathnayake

Guest

Applied Mathematics [linear motion :( I guess)]

A train moves with constant velocity u m/s. but one day due to a railway repair, it deaccelerated with (constant retardation) f and achieved rest. but immediately it started to accelerate with (constant acceleration) 2f and became to it's constant velocity u m/s.  use the v-t graph to prove the delay due to repair is 3u/4f.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: Applied Mathematics [linear motion :( I guess)]

hi Thimira

To make the graph draw a line across for the time axis and a line up for the velocity axis.  The gradient of any line representing a change in velocity is what we call the acceleration.

So start at u on the up axis and draw a (straight sloping) line down to the across axis.  This line has gradient -f (minus as it's decelerating).

Now draw a line sloping up and stop when it reaches the same height (u) as before.  As this acceleration is 2f make this line look twice as steep.  As it's only a sketch you don't have to measure this exactly.  So the V-T graph looks a bit like a V but part two of the V is steeper than part one.

Gradient is up/across so across = up/gradient.

For the first part the time taken = across amount = u/f

For the second part the time taken is u/2f.

So the total time taken is u/f + u/2f = 2u/2f + u/2f = 3u/2f

If the train doesn't slow down at all then its V-T graph is a horizontal line at height u.

The distance travelled is found by calculating the area under the graph.  For the decelerating/accelerating graph that's the area of the two triangular sections.  If the train is going at a constant speed then its distance is  a rectangular section.  For the same height shapes the rectangle only needs to be half as long, ie. 3u/4f and the time lost due to slowing is the same amount 3u/4f.

Bob

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