A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.
Been stuck on it for atleast 2 hours, please help!

Raeve
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 15012017 16:42

Revision help in partnership with Birmingham City University

RDKGames
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 15012017 16:49
(Original post by Raeve)
A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.
Been stuck on it for atleast 2 hours, please help!
Particle 1:
Particle 2:
Construct they will overtake eachother when their displacements are equal.
Two equations in and and solve them. 
Mr M
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 15012017 16:51
(Original post by Raeve)
A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.
Been stuck on it for atleast 2 hours, please help! 
Raeve
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 15012017 16:53
(Original post by RDKGames)
2 hours? lol damn...
Particle 1:
Particle 2:
Construct they will overtake eachother when their displacements are equal.
Two equations in and and solve them.
s = 1/2*3.2*t^2
s=3t+2.7*t^2
I then equated them, but never got the correct answer. 
Raeve
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 15012017 16:54
(Original post by Mr M)
Hi. The best place to post A level maths questions is F38 (the Maths Forum), 
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 15012017 16:59
(Original post by Raeve)
Oh sorry, it is my first post. I did not realise. 
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 15012017 17:01
(Original post by Mr M)
No problem. I've asked for the thread to be moved. In the meantime, you are in good hands with RDKGames. 
Student1256
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 15012017 17:02
you have to make two simultaneous equations and solve them. let us start off by imagining the instant the second ball has arrived at O.
what is the distance of first particle from O? s = (5)^2 * 3.2/2 = 25*1.6
what is the current speed of particle one? u = at = 5*3.2 = 16
now since we want to know the time both particles are the same distance from O we equate the equations of distance from point O
eq of first particle = 25*1.6 + 16t + (3.2t^2)/2
eq of second particle = 3t + (5.4t^2)/2
the t you will get by solving the quadratic equation you end up with is actually 5 seconds after the t you want because we started from the instant the second ball was at O to simplify things. so just minus 5 from the t you end up with.
after you get the t use the first t and input it into either of the two equations and get the distance from O.
Hope this helped. 
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 15012017 17:05
(Original post by RDKGames)
2 hours? lol damn...
Particle 1:
Particle 2:
Construct they will overtake eachother when their displacements are equal.
Two equations in and and solve them. 
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 15012017 17:09
(Original post by Raeve)
Oh, good to know, thanks! 
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 15012017 17:09
(Original post by Student1256)
you have to make two simultaneous equations and solve them. let us start off by imagining the instant the second ball has arrived at O.
what is the distance of first particle from O? s = (5)^2 * 3.2/2 = 25*1.6
what is the current speed of particle one? u = at = 5*3.2 = 16
now since we want to know the time both particles are the same distance from O we equate the equations of distance from point O
eq of first particle = 25*1.6 + 16t + (3.2t^2)/2
eq of second particle = 3t + (5.4t^2)/2
the t you will get by solving the quadratic equation you end up with is actually 5 seconds after the t you want because we started from the instant the second ball was at O to simplify things. so just minus 5 from the t you end up with.
after you get the t use the first t and input it into either of the two equations and get the distance from O.
Hope this helped. 
RDKGames
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 15012017 17:12
(Original post by Raeve)
I tried that:
s = 1/2*3.2*t^2
s=3t+2.7*t^2
I then equated them, but never got the correct answer.
It's the fact that the second particle comes in 5 seconds later that's the key you are missing.
Begin by finding the distance travelled by P1 from the origin within the first 5 seconds, and the speed, then equate the distance travelled by P1 in the first 5 seconds + (the displacement of P1 from 5s onwards) with the displacement of the second particle.
Then just adapt my method above. 
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 15012017 17:12
(Original post by Raeve)
Oh that is great thanks, so I was pretty much there I just didn't add the 40m onto the equation. Infact why do you add it, because we are using, s=ut+1/2at^2? 
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 15012017 17:16
(Original post by Student1256)
because that is the distance the particle is already away from O and then we are assuming t = 0 at that instant in time when it's that far from O. if you still dont get it basically s = ut + 0.5at^2 means the distance away from a point when you are at that point at t = 0. but when you are already 40 m away from that point at t = 0 you just add it into the equation. 
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 15012017 17:18
(Original post by Student1256)
because that is the distance the particle is already away from O and then we are assuming t = 0 at that instant in time when it's that far from O. if you still dont get it basically s = ut + 0.5at^2 means the distance away from a point when you are at that point at t = 0. but when you are already 40 m away from that point at t = 0 you just add it into the equation. 
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 15012017 17:19
(Original post by Raeve)
Oh of course! Thanks a lot. I appreciate the help. 
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 15012017 17:21
(Original post by Raeve)
A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.
Been stuck on it for atleast 2 hours, please help!
That gives you s = 1.6t^2 and s = 3(t5) + 2.7(t5)^2 
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 15012017 17:22
(Original post by Student1256)
that is wrong because you assume the second particle is accelerating since t = 0 while that is wrong and you cant form any normal single equation to describe the motion of the second ball as it starts accelerating after a certain time. its best to use my method of considering the two balls at the instant in time that the second particle starts to accelerate and has reached O
(Original post by Raeve)
Oh of course! Thanks a lot. I appreciate the help.
which is particle 1 in order to compensate for the late start of particle 2
and then just solve them. 
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 15012017 17:23
(Original post by Raeve)
A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.
Been stuck on it for atleast 2 hours, please help!
If you say 't' is the time from when the second particle passes through O then at t = 0, the first particle will have been travelling for 5 seconds i.e. t + 5. So you get
Second particle:
u= 3
a = 5.4
t = t
First particle:
u = 0
a = 3.2
t = t + 5
Then you can equate the two displacements.
EDIT : Well it hadn't been suggested before I started writing this postLast edited by Notnek; 15012017 at 17:25. 
Raeve
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 15012017 17:24
(Original post by RDKGames)
Yeah misread.
Alternatively, to skip some little working, you could've had:
which is particle 1 in order to compensate for the late start of particle 2
and then just solve them.
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