M1 Question help

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Macceroot
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#1
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#1
There are two questions in which I would be very grateful if I could receive help from relationg to kinematics of a particle moving in a straight line:

(1) A particle is projected vertically upwards from a point O with speed u ms^-1. Two seconds later it is still moving upwards and its speed is 1/3u ms^-1. Find a the value of u, b the time from the instant that the particle leaves O to the instant that it returns to O

(2) A Particle P is projected vertically upwards from a point O with speed 12ms^-1. One second after P has been projected from O, another particle Q is projected vertically upwards from O with speed 20ms^-1. Find a the time between the instant that P is projected from O and the instant when P and Q collide, b the distance of the point where P and Q collide from O
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graham11238
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#2
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#2
ok now it does
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graham11238
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#3
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ok for 2., use equation s=ut+(0.5)at^2, for both particles s (distance) will be the same at collison and time of collision let us say it is T for particle P and T-1 for particle Q. as T is the only unknown it can be found
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graham11238
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#4
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also for one use this equation to find U
1/3U=U-(9.8)(2)

the rest is pretty straight forward
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Macceroot
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#5
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(Original post by graham11238)
ok for 2., use equation s=ut+(0.5)at^2, for both particles s (distance) will be the same at collison and time of collision let us say it is T for particle P and T-1 for particle Q. as T is the only unknown it can be found
I managed to work out the first one, thanks

For the second, I got to this stage:

P: a=-9.8 u=12 s=12t-9.8t^2

Q: a=-9.8 u=20 s=20t-9.8t^2

I don't understand the next part as the second difference confuses me..
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graham11238
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#6
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set: s=12t-9.8t^2 equal to s=20(t-1)-9.8(t-1)^2

the whole bracket on the second eqn is squared
so
12t-9.8t^2=20(t-1)-9.8(t-1)^2

solve as a linear as the t^2 cancels out
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Macceroot
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#7
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(Original post by graham11238)
set: s=12t-9.8t^2 equal to s=20(t-1)-9.8(t-1)^2

the whole bracket on the second eqn is squared
so
12t-9.8t^2=20(t-1)-9.8(t-1)^2

solve as a linear as the t^2 cancels out
Ok I get that thanks, but I come to the wrong answer still..

I get 12t-9.8t^2=20t-20-9.8t^2+19.6t-9.8

Eventually making 27.6t=29.8 t=1.1 seconds.

The answer in my book says 1.4 seconds.

Is that an error?
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graham11238
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#8
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sorry, let me work it out for myself, i'll get back to you
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graham11238
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#9
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i get t as 1.465 is that right or is it exactly 1.4?
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graham11238
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#10
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#10
sorry not sure i can help
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Macceroot
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#11
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(Original post by graham11238)
i get t as 1.465 is that right or is it exactly 1.4?
No you are correct, the answer is 1.4 to 2 significant figures ( you obviously rounded it up as some point or they did)

What have I done wrong in my working to get 1.1?


By the way thanks for all this help
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graham11238
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#12
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(Original post by Macceroot)
No you are correct, the answer is 1.4 to 2 significant figures ( you obviously rounded it up as some point or they did)

What have I done wrong in my working to get 1.1?


By the way thanks for all this help
hmm not sure, i thought that the equation s=ut+(0.5)at^2 applies to both where to give time a vaule of say T for particle P and (T-1) for particle Q as Q sets off one second later. working this through you get an equation

12(T)-(4.9)(t^2)=20(T-1)-(4.9)(T-1)^2

12T=20T-20+9T-4.9

24.9=17T

1.46470=T

This should work out exactly but when i plug the numbers back in it seems to give different values of s for P and Q which doesn't make sense.

what do you think
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DeltaGiang
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#13
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#13
1.4 seconds is correct. I found it equals approximately 1.3988s
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davros
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#14
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#14
(Original post by DeltaGiang)
1.4 seconds is correct. I found it equals approximately 1.3988s
Please check the dates of threads before you reply - this question is 12 years old!
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