# Thread: SUVAT - Motion Equation

1. ## SUVAT - Motion Equation

This has something to do with graphing (speed time graph) so I suspect this is the right forum.

Question: Two stations A and B are 495m apart. A train starts from A and accelerates uniformly for 15s to a speed of $\displaystyle 54kmh^{-1}$ which it maintains until it is 90m from B. Sketch a speed-time graph to show the motion of the train. Find the magnitude of the acceleration and retardation and the total time the journey takes.

I'm stuck on the magnitude of acceleration =\ .

2. Originally Posted by Viral
This has something to do with graphing (speed time graph) so I suspect this is the right forum.

Question: Two stations A and B are 495m apart. A train starts from A and accelerates uniformly for 15s to a speed of $\displaystyle 54kmh^{-1}$ which it maintains until it is 90m from B. Sketch a speed-time graph to show the motion of the train. Find the magnitude of the acceleration and retardation and the total time the journey takes.

I'm stuck on the magnitude of acceleration =\ .

HI

Have you tried $\displaystyle s=ut+\frac{1}{2}at^2$ ?

3. I'm not sure what it means by magnitude of acceleration. Because when we've seen magnitude before, we used pythagoras to get the hypotenuse of the triangle for the magnitude >< .

4. Originally Posted by Viral
I'm not sure what it means by magnitude of acceleration. Because when we've seen magnitude before, we used pythagoras to get the hypotenuse of the triangle for the magnitude >< .
It means value , in this case value of the acceleration , just that this value is never negative . So use the formula i gave you to find its magnitude .

5. Ahh, so it's always the absolute value of the acceleration?

EDIT: So I find the magnitude of the acceleration, then the magnitude of the retardation and find the difference?

6. Originally Posted by Viral
Ahh, so it's always the absolute value of the acceleration?

EDIT: So I find the magnitude of the acceleration, then the magnitude of the retardation and find the difference?
yes .

Not quite true . THe train accelerates until a point 90 m before B . Then from here , the train decelerates until station B .

So for acceleration , you know s=405 m , u=54 km/h (convert this to m/s) , and t=15 s

so by s=ut+1/2at^2 , you can find a .

Now for deceleration ,

s=90 , u=54 km/h (again , do the conversion) , v=0

use $\displaystyle v^2=u^2+2as$ , you should get a negative value for a since its deceleration.

7. I realised that I did not have the displacement from the first motion of acceleration. Therefore I used $\displaystyle v=u+at$ and worked out the right answer of $\displaystyle 1ms^{-1}$. I successfully worked out the retardation as $\displaystyle 1.25ms^{-1}$.

8. Originally Posted by Viral
I realised that I did not have the displacement from the first motion of acceleration. Therefore I used $\displaystyle v=u+at$ and worked out the right answer of $\displaystyle 1ms^{-1}$. I successfully worked out the retardation as $\displaystyle 1.25ms^{-1}$.
Are you sure those are the answers given ? by the way , your unit is wrong too .

9. yep they're the right answers (and those were typos, they should be $\displaystyle ms^{-2}$.