# SUVAT - Motion Equation

• Oct 11th 2009, 06:29 AM
Viral
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 =\ .
• Oct 11th 2009, 06:37 AM
Quote:

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 $\displaystyle s=ut+\frac{1}{2}at^2$ ?
• Oct 11th 2009, 06:39 AM
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 >< .
• Oct 11th 2009, 06:42 AM
Quote:

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 .
• Oct 11th 2009, 06:43 AM
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?
• Oct 11th 2009, 06:53 AM
Quote:

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.
• Oct 11th 2009, 07:41 AM
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 $\displaystyle 1.25ms^{-1}$.
• Oct 11th 2009, 07:49 AM
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 $\displaystyle 1.25ms^{-1}$.
yep they're the right answers (and those were typos, they should be $\displaystyle ms^{-2}$.