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Macceroot
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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
(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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graham11238
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#3
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
also for one use this equation to find U
1/3U=U-(9.8)(2)
the rest is pretty straight forward
1/3U=U-(9.8)(2)
the rest is pretty straight forward
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Macceroot
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#5
(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
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
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
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
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
(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
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
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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graham11238
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Macceroot
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#11
(Original post by graham11238)
i get t as 1.465 is that right or is it exactly 1.4?
i get t as 1.465 is that right or is it exactly 1.4?
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
(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
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
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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davros
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#14
(Original post by DeltaGiang)
1.4 seconds is correct. I found it equals approximately 1.3988s
1.4 seconds is correct. I found it equals approximately 1.3988s
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