# Thread: Motion with uniform acceleration??

1. ## Motion with uniform acceleration??

Hi, my first post dont know if im on the right one :S.

Can someone help with this maths question, currently using SUVAT method.

A train leaves a station from rest with a constant acceleration of 0.3ms-1. It reaches its maximum speed after 90 seconds and mantains this speed for a futher 5 minutes, when it slows down to a stop at a second station with an acceleration of -1ms-2. How far apart are the two station.

Ive figured that the first 90 seconds its travelled 1215m I think. Can anyone help me with the rest. If you could to explain how you did it. I want to understand. Thankyou.

2. Originally Posted by NathanBUK Hi, my first post dont know if im on the right one :S.

Can someone help with this maths question, currently using SUVAT method.

A train leaves a station from rest with a constant acceleration of 0.3ms-1. It reaches its maximum speed after 90 seconds and mantains this speed for a futher 5 minutes, when it slows down to a stop at a second station with an acceleration of -1ms-2. How far apart are the two station.

Ive figured that the first 90 seconds its travelled 1215m I think. Can anyone help me with the rest. If you could to explain how you did it. I want to understand. Thankyou.
By the "SUVAT method" I presume that's some acronym that somehow encapsulates $\displaystyle v = u + at, s = ut+\frac 1 2 a t^2, v^2 = u^2 + 2 a s$.

Split the thing up into 3 bits. I'm leaving out the units (metres, seconds, etc.) for clarity, but be aware your teacher will crucify you if you do the same.

1. You have a = 0.3, u = 0, t = 90. From that you can easily calculate the distance by $\displaystyle s = ut + \frac 1 2 a t^2$ (which you did, whether you got the right answer depends on how accurately you punched your calculator keys). Now, what you *also* want to do is get the velocity at this point - so use $\displaystyle v = u + at$ for that (and you know u = 0 and a = what I said above.

2. Now you know how fast it's going, and you've got a = 0 so $\displaystyle s = ut (+ \frac 1 2 a t^2)$ and that gives you the second distance.

3. In this case, you have $\displaystyle v^2 = u^2 + 2 a s$ again, but this time $\displaystyle a = -1$, u is the velocity from the last bit, and $\displaystyle v = 0$. Then rearrange the equation so it's in terms of s.

Add them up together and you're done.

3. Oh Thankyou soo much!! Just starting out on this and well homework and no teacher come a bit stuck but thankyou so much!!

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