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# M1 Maths HELP

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1. A particle accelerates from rest with an acceleration of 3ms^-2 to speed V. It continues at this speed for time T then decelerates to rest at 1.5ms ^-2.
The total time for this motion is 60 seconds and the total distance travelled is 1000 m, find the value of V.

I've tried so many method and they're not working
2. (Original post by jordanshelley97)
You can use SUVAT.

S = 1000
U = 0
V = ?
A = 3
T = ?

We can use v^2 = u^2 + 2as
v^2 = 6000
v = square root of 6000
Isn't that incorrect because it doesn't have an acceleration of 3ms^-2 throughout the whole motion
3. (Original post by Jozanic)
Isn't that incorrect because it doesn't have an acceleration of 3ms^-2 throughout the whole motion
Draw a speed-time graph and it can help alongside SUVAT. The graphs has 3 parts where you can apply SUVAT - firstly when it's accelerating at 3m/s^2, then when it's not accelerating, and finally when it is decelerating.

The area under this graph should be 1000m.
4. (Original post by jordanshelley97)
You can use SUVAT.

S = 1000
U = 0
V = ?
A = 3
T = ?

We can use v^2 = u^2 + 2as
v^2 = 6000
v = square root of 6000
Acceleration isn't constant throughout the entire 1000m therefore it doesn't work the way you did it.
5. (Original post by RDKGames)
Draw a speed-time graph and it can help alongside SUVAT. The graphs has 3 parts where you can apply SUVAT - firstly when it's accelerating at 3m/s^2, then when it's not accelerating, and finally when it is decelerating.

The area under this graph should be 1000m.
I did exactly that but I'm still getting weird answers.
6. (Original post by Jozanic)
I did exactly that but I'm still getting weird answers.
7. You could draw a speed time graph. The area under the graph is the total distance travelled.
8. A Speed time graph can be drawn here.
9. (Original post by RDKGames)
AB (Accelerating)
s =
u = 0
v = V
a = 3
t =

BC (Constant)
s =
u = V
v = V
a = 0
t = T

CD (Decelerating)
s =
u = V
v = 0
a = -1.5
t =

I made S the subject using v^2 = u^2+2as and tried to equate the accelerating and decelerating part.
10. (Original post by Jozanic)
AB (Accelerating)
s =
u = 0
v = V
a = 3
t =

BC (Constant)
s =
u = V
v = V
a = 0
t = T

CD (Decelerating)
s =
u = V
v = 0
a = -1.5
t =

I made S the subject using v^2 = u^2+2as and tried to equate the accelerating and decelerating part.
Try the following:

and and

You can easily get this from drawing the graph and labeling the end of first interval as , end of second as and end of third as 60.
11. (Original post by RDKGames)
Try the following:

and and

You can easily get this from drawing the graph and labeling the end of first interval as , end of second as and end of third as 60.
Ahhhh thanks. I sort of get the idea now. You're adding the separate parts of the graph. But what is the second section with V/t1 = 3 and -V/(60/t2) =-1.5
12. (Original post by Jozanic)
Ahhhh thanks. I sort of get the idea now. You're adding the separate parts of the graph. But what is the second section with V/t1 = 3 and -V/(60/t2) =-1.5
Gradients of a speed-time graph are the acceleration, I'm merely putting the acceleration's in terms of V and t.
13. (Original post by RDKGames)
Gradients of a speed-time graph are the acceleration, I'm merely putting the acceleration's in terms of V and t.
Ahhh I see. Thanks a bunch
14. (Original post by RDKGames)
Gradients of a speed-time graph are the acceleration, I'm merely putting the acceleration's in terms of V and t.
Ok I'm slowly getting an understanding but what do you do after ???#

EDIT: Actually never mind I solved it.
15. (Original post by Jozanic)
Ok I'm slowly getting an understanding but what do you do after ???#

EDIT: Actually never mind I solved it.
Rearrange for and in terms of then sub them into the formula to get a quadratic in
16. (Original post by Uni12345678)
I got this
Hi, appreciate you trying to help but please do not post full solutions as it's against the forum's rules. Thanks.

http://www.thestudentroom.co.uk/show...9#post64637319
17. (Original post by RDKGames)
Hi, appreciate you trying to help but please do not post full solutions as it's against the forum's rules. Thanks.

http://www.thestudentroom.co.uk/show...9#post64637319
Hi again. I got the quadratic 4v^2 - 181 + 3000 = 0

Am I on the right lines?
18. (Original post by Jozanic)
Hi again. I got the quadratic 4v^2 - 181 + 3000 = 0

Am I on the right lines?
No, that would give you imaginary roots for the speed. Try again and be careful with your algebraic manipulation.

You should have and
19. (Original post by RDKGames)
Hi, appreciate you trying to help but please do not post full solutions as it's against the forum's rules. Thanks.

http://www.thestudentroom.co.uk/show...9#post64637319
Fair soz
20. (Original post by RDKGames)
No, that would give you imaginary roots for the speed. Try again and be careful with your algebraic manipulation.

You should have and
Still getting the same equation

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