# Maths: Kinematics

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Thread starter 1 month ago
#1
Please help with this question: A particle is projected at a speed of 12ms-1 up a straight inclined track. Whilst on the track the particle experiences a constant acceleration down the track of 5ms-2. Find the length of time that the particle is more than 8 metres from the point of projection.
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1 month ago
#2
not sure how to do the whole question but to start use the SUVAT equations ->
s = 8
u = 12
v = v
a = 5
t = ?
Last edited by milliebaudot; 1 month ago
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1 month ago
#3
(Original post by Risermax)
Please help with this question: A particle is projected at a speed of 12ms-1 up a straight inclined track. Whilst on the track the particle experiences a constant acceleration down the track of 5ms-2. Find the length of time that the particle is more than 8 metres from the point of projection.
Do you have an answer for this?
(I have one but not sure it’s right)
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Thread starter 1 month ago
#4
(Original post by GabiAbi84)
Do you have an answer for this?
(I have one but not sure it’s right)
No I do not
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1 month ago
#5
(Original post by Risermax)
No I do not
Ah, shoot. I like to know I’m right before I help..

What have you done so far?
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1 month ago
#6
(Original post by milliebaudot)
not sure if this helps as I'm not sure how to do the whole question but to start use the SUVAT equations ->
s = 8
u = 12
v = v
a = 5
t = ?
Please just post hints - it's against forum rules to post more than that.
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Thread starter 1 month ago
#7
(Original post by GabiAbi84)
Ah, shoot. I like to know I’m right before I help..

What have you done so far?
I just wrote the suvat for it but Im still not sure whether gravity has anything to do with it so is a: 5 or 9.81
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Thread starter 1 month ago
#8
(Original post by Risermax)
I just wrote the suvat for it but Im still not sure whether gravity has anything to do with it so is a: 5 or 9.81
-9.81*
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1 month ago
#9
(Original post by Risermax)
I just wrote the suvat for it but Im still not sure whether gravity has anything to do with it so is a: 5 or 9.81
It is on a track and the question tells you the constant acceleration is 5m/s DOWN the track.
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Thread starter 1 month ago
#10
(Original post by GabiAbi84)
It is on a track and the question tells you the constant acceleration is 5m/s DOWN the track.
So did you get 0.44 (3sf) as your final answer
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1 month ago
#11
(Original post by Risermax)
So did you get 0.44 (3sf) as your final answer
I did not...but that doesn’t mean you’re not correct. Show your working
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Thread starter 1 month ago
#12
(Original post by GabiAbi84)
I did not...but that doesn’t mean you’re not correct. Show your working
I did s = ut+1/2at^2
Subbed in the values to get
8 = 12t+2.5t^2
Then did 2.5t^2+12t-8 = 0
Did the quadratic formula and got 0.5933...
So not 0.44 my bad
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1 month ago
#13
(Original post by Risermax)
I did s = ut+1/2at^2
Subbed in the values to get
8 = 12t+2.5t^2
Then did 2.5t^2+12t-8 = 0
Did the quadratic formula and got 0.5933...
So not 0.44 my bad
You’re looking for the length of time it is More than 8 metres away.
Also your acceleration is down the track so should be negative.

(My confusion lies in whether it returns back to the start once it gets to the top point of the track)
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1 month ago
#14
(Original post by GabiAbi84)
You’re looking for the length of time it is More than 8 metres away.
Also your acceleration is down the track so should be negative.

(My confusion lies in whether it returns back to the start once it gets to the top point of the track)
It should do. The acceleration will remain.
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1 month ago
#15
(Original post by Retread)
It should do. The acceleration will remain.
Thanks
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Thread starter 1 month ago
#16
(Original post by GabiAbi84)
You’re looking for the length of time it is More than 8 metres away.
Also your acceleration is down the track so should be negative.

(My confusion lies in whether it returns back to the start once it gets to the top point of the track)
I don't think it does, but how did you work it out?
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1 month ago
#17
(Original post by Risermax)
I don't think it does, but how did you work it out?
I started by working out
Last edited by GabiAbi84; 1 month ago
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Thread starter 1 month ago
#18
(Original post by GabiAbi84)
I started by working out the time until it stops moving upwards.
Please tell me more
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1 month ago
#19
(Original post by Risermax)
Please tell me more
Have you done that?
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Thread starter 1 month ago
#20
(Original post by GabiAbi84)
Have you done that?
Yes and got 0.
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