# Thread: Another speed time graph problem

1. ## Another speed time graph problem

A train starts from rest and moves with uniform acceleration for 5 minutes. It then maintains a constant speed for 20 minutes before being brought to rest by a uniform retardation of magnitude twice that of the acceleration. Sketch a speed time graph of the motion and find the time it takes the train to stop. If the train travels 4.5 km while accelerating, show that the acceleration is $\displaystyle \frac{1}{10}ms^{-2}$ and find the total distance travelled.

Not sure how to work out how long it takes for the train to stop.

2. Originally Posted by Viral
A train starts from rest and moves with uniform acceleration for 5 minutes. It then maintains a constant speed for 20 minutes before being brought to rest by a uniform retardation of magnitude twice that of the acceleration. Sketch a speed time graph of the motion and find the time it takes the train to stop. If the train travels 4.5 km while accelerating, show that the acceleration is $\displaystyle \frac{1}{10}ms^{-2}$ and find the total distance travelled.

Not sure how to work out how long it takes for the train to stop.
2.5 minutes to stop ... why?

3. That's what I'm asking, I don't know how to work it out.

4. Originally Posted by Viral
A train starts from rest and moves with uniform acceleration for 5 minutes. Sketch a speed time graph of the motion and find the time it takes the train to stop. If the train travels 4.5 km while accelerating, show that the acceleration is $\displaystyle \frac{1}{10}ms^{-2}$ and find the total distance travelled.

Not sure how to work out how long it takes for the train to stop.
Have you drawn the graph yet? The time it takes for the train to stop will be where that graph crosses the t axis.
A train starts from rest and moves with uniform acceleration for 5 minutes.
So the graph is a straight line from (0,0) to (5, 5a) where a is the (as yet unknown) acceleration in km/min.

It then maintains a constant speed for 20 minutes
The graph is a horizontal line from (5, 5a) to (20, 5a). You still don't know a.

before being brought to rest by a uniform retardation of magnitude twice that of the acceleration.
The acceleration is now -2a. Since it now starts with a speed of 5a its speed after t minutes will be 5a- 2at and that will be 0 when 2at= 5a or then t= 5/2. That's reasonble- at twice the deceleration, the train will stop in half the time it took to accelerate. The clocks stops at 2+ 5/2= 22.5 minutes after starting. The graph is a straight line from (20, 5a) to (22.5, 0).

That graph should look like a "trapezoid" with lower base of length $\displaystyle b_1= 22.5- 0= 22.5$ and upper base of length $\displaystyle b_2= 20- 5= 15$. The height is $\displaystyle h= 5a$.

Now- the total distance traveled is the area of that trapezoid! The area of a trapezoid is given by $\displaystyle h\frac{b_1+ b_2}{2}$ which, here, is $\displaystyle 5a\frac{22.5+ 15}{2}= 93.75a$ km.

Unfortunately, that depends upon that still unknown acceleration, a.

Fortunately, we are told that, during the initial 5 min. acceleration, the train moved 4.5 km. If we stop the graph at t= 5, we have a triangle with base of length b= 5 and height h= 5a. The area of that triangle is this first distance. The area of the triangle is $\displaystyle \frac{1}{2}bh= \frac{25a}{2}= 4.5$. You can solve that for a.

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### v at 6m/sfor 5min accelerat 24m/s in20 min speed for two min and halting 8s calculate acc and deceleration draw the graph represent

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