# kinematics problem

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• Nov 9th 2008, 11:31 AM
ellie_h
kinematics problem
i have like 5 questions the same style and i cant do them, so if somebody could work through this 1 ill be really grateful.

Two trains A and B run on parallel striaght tracks. Initially both are at rest in the station and level with each other. At time t=0, A starts to move. It moves with constant acceleration for 12 sec up to a speed of 30m/s, and then moves at a constant speed of 30m/s. Train B starts to move in the same direction as A when t=40, where t is measure in seconds. It accelerates with the same initial acceleration as A, up to a speed of 60m/s. it then moves at a constant speed of 60m/s. Train B overtakes A after both have reached their maximum speed. Train B overtakes A when t = T.

We had to draw the speed time graphs of both trains for 0 ≤ t ≤ T.
The second part is find the value of T.
• Nov 9th 2008, 12:30 PM
skeeter
hopefully, you know why it helps to draw a velocity vs. time graph.

the area between the graph and the time axis is displacement.

displacement of train A = $\frac{1}{2}(12)(30) + 30(T-12)$

displacement of train B = $\frac{1}{2}(24)(60) + 60(T-64)$

the two trains have the same displacement when B passes A.
• Nov 9th 2008, 12:45 PM
ellie_h
Quote:

Originally Posted by ellie_h
i have like 5 questions the same style and i cant do them, so if somebody could work through this 1 ill be really grateful.

Two trains A and B run on parallel striaght tracks. Initially both are at rest in the station and level with each other. At time t=0, A starts to move. It moves with constant acceleration for 12 sec up to a speed of 30m/s, and then moves at a constant speed of 30m/s. Train B starts to move in the same direction as A when t=40, where t is measure in seconds. It accelerates with the same initial acceleration as A, up to a speed of 60m/s. it then moves at a constant speed of 60m/s. Train B overtakes A after both have reached their maximum speed. Train B overtakes A when t = T.

We had to draw the speed time graphs of both trains for 0 ≤ t ≤ T.
The second part is find the value of T.

Quote:

Originally Posted by skeeter
hopefully, you know why it helps to draw a velocity vs. time graph.

the area between the graph and the time axis is displacement.

displacement of train A = \frac{1}{2}(12)(30) + 30(T-12)

displacement of train B = \frac{1}{2}(24)(60) + 60(T-64)

the two trains have the same displacement when B passes A.

so then do i have 2 simultanious equations to solve or something like that?
• Nov 9th 2008, 12:53 PM
skeeter
all set up for you ...

displacement of train A = $\frac{1}{2}(12)(30) + 30(T-12)$

displacement of train B = $\frac{1}{2}(24)(60) + 60(T-64)$

displacement of train A = displacement of train B

set the two equal and solve for T.
• Nov 9th 2008, 01:10 PM
ellie_h
Quote:

Originally Posted by skeeter
all set up for you ...

displacement of train A = $\frac{1}{2}(12)(30) + 30(T-12)$

displacement of train B = $\frac{1}{2}(24)(60) + 60(T-64)$

displacement of train A = displacement of train B

set the two equal and solve for T.

i know that i thught i understood but i actually dont see how you got the two equations.

you see i missed school an want to catch up, but i dont understand how to do these problems, so i thought if somebody could explain it exactly ill be able to do the other problems.

i know that this is really rude but do you mind explaining it in a bit more depth?