You are Here: Home >< Maths

# M1 - SUVAT help watch

1. I need help with this question.

A particle is moving along a straight line OA with constant acceleration 2ms-2. At O the particle is moving towards A with speed 5.5ms-1. The distance OA is 20m. Find the time the particle takes to move from O to A.

S=20 V=5.5 A=2 T=?

I used the formula s=vt-1/2at^2 but I got the wrong answer.

Any help?
2. How can the particle be moving at 5.5ms-1 at OA? Since OA is a distance over which it has non-zero acceleration. Are you sure it's not just at O?
3. Well, all I can suggest is that you try again. I have an answer; what did you get and what should the answer be, before I say?
4. v is not 5.5, its 'u'
5. (Original post by ForGreatJustice)
How can the particle be moving at 5.5ms-1 at OA? Since OA is a distance over which it has non-zero acceleration. Are you sure it's not just at O?
I'm assuming it means the speed at A is 5.5ms-1
6. (Original post by ForGreatJustice)
How can the particle be moving at 5.5ms-1 at OA? Since OA is a distance over which it has non-zero acceleration. Are you sure it's not just at O?
That was a typo, i've edited the question.

(Original post by DarkWhite)
I have an answer; what did you get and what should the answer be, before I say?
Spoiler:
Show
2.5
7. (Original post by Mist786)
That was a typo, i've edited the question.

Spoiler:
Show
2.5
Ok, in which case remember that v is the final velocity, but you're only given the initial velocity u, for which there is a (very similar) formula for which will work
8. (Original post by ForGreatJustice)
Ok, in which case remember that v is the final velocity, but you're only given the initial velocity u, for which there is a (very similar) formula for which will work
I still get the wrong answer when I substitute the values, I think i'm rearrange it wrong. Any help? I'll rep
9. s=ut+1/2at^2 then if you substitute the values in, you get a quadratic in t, which gives the answer.

If you still get the wrong anwser, post your working.
10. (Original post by ForGreatJustice)
s=ut+1/2at^2 then if you substitute the values in, you get a quadratic in t, which gives the answer.

If you still get the wrong anwser, post your working.
20=5.5t-1/2*2*t^2
20=5.5t-t^2
20=t(5.5+t)
20/t=5.5+t

^^That's wrong .. idk
11. ok...you're using the wrong equation, we want s=ut + 1/2at^2

putting the numbers in to this gives 20=5.5t+t^2

t^2+5.5t-20=0

which can be either factorised (easiest if you can spot it), completing the square (messy in this case) or the quadratic formula.

You will get 2 values for t, but clearly one will be absurd in this context, and the other is the answer you are looking for.
12. You need to use two of the formulae - here goes.

S= 20 U = 5.5 V = ? A= 2 T= ?

V^2=U^2+2AS
Therefore V = Sqrt(30.25+80) = 10.5

Then plug into V= U+AT (Which we'll rearrange to (V-U)/A = T ) and you have (10.5-5.5)/2 = 2.5 = T
13. (Original post by Longorefisher)
You need to use two of the formulae - here goes.

S= 20 U = 5.5 V = ? A= 2 T= ?

V^2=U^2+2AS
Therefore V = Sqrt(30.25+80) = 10.5

Then plug into V= U+AT (Which we'll rearrange to (V-U)/A = T ) and you have (10.5-5.5)/2 = 2.5 = T
you can, but the one simple formula works.
14. (Original post by ForGreatJustice)
ok...you're using the wrong equation, we want s=ut + 1/2at^2

putting the numbers in to this gives 20=5.5t+t^2

t^2+5.5t-20=0

which can be either factorised (easiest if you can spot it), completing the square (messy in this case) or the quadratic formula.

You will get 2 values for t, but clearly one will be absurd in this context, and the other is the answer you are looking for.
I still get the wrong answer when I do the quadratic formula and I have no idea how to do completing the square.
15. (Original post by ForGreatJustice)
you can, but the one simple formula works.
Yeah, I would agree although I just ran the quadratic formula on it and it's not coming out at the answer I'd expect.
Edit - calc error. Using the wrong co-efficent of x^2 - I do agree with you now.
16. you shouldn't do, but the quadratic factorises to (t-2.5)(t+8)=0 which gives the answer

just tried the quadratic formula, that works too

out of curiosity what did you get from the quadratic formula?
17. (Original post by ForGreatJustice)
you shouldn't do, but the quadratic factorises to (t-2.5)(t+8)=0 which gives the answer

just tried the quadratic formula, that works too

out of curiosity what did you get from the quadratic formula?
The quadratic formula gave me -45.5 and -40

Don't laugh at me

18. where

So you just put a=1, b=5.5, and c=-20 in

are you just putting the whole load into a calculator? since if you don't bracket up right, things go wrong
19. (Original post by ForGreatJustice)

where

So you just put a=1, b=5.5, and c=-20 in

are you just putting the whole load into a calculator? since if you don't bracket up right, things go wrong
lol I realised my mistake

Thanks again

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 4, 2010
The home of Results and Clearing

### 1,942

people online now

### 1,567,000

students helped last year
Today on TSR

### Took GCSEs this summer?

Fill in our short survey for Amazon vouchers!

### University open days

1. Bournemouth University
Wed, 22 Aug '18
2. University of Buckingham
Thu, 23 Aug '18
3. University of Glasgow
Tue, 28 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams