# M1 Kinematics

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#1
A particle X, moving along a straight line with constant speed 4 m s − 1, passes through a fixed point O. Two seconds later, another
particle Y, moving along the same straight line and in the same direction, passes through O with speed 6 m s − 1. The particle Y is
moving with constant deceleration 2 m s − 2
a Write down expressions for the velocity and displacement of each particle t seconds after Y passed through O.
b Find the shortest distance between the particles after they have both passed through O.
c Find the value of t when the distance between the particles has increased to 23 m.

part b is the one i'm really kind of struggling with, i don't understand the ms's method to approach the answer.
Last edited by Aleksander Krol; 1 month ago
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#2
(Original post by Aleksander Krol)
A particle X, moving along a straight line with constant speed 4 m s − 1, passes through a fixed point O. Two seconds later, another
particle Y, moving along the same straight line and in the same direction, passes through O with speed 6 m s − 1. The particle Y is
moving with constant deceleration 2 m s − 2
a Write down expressions for the velocity and displacement of each particle t seconds after Y passed through O.
b Find the shortest distance between the particles after they have both passed through O.
c Find the value of t when the distance between the particles has increased to 23 m.
for part A: i did it as:
Ans: Particle X: u=4 m/s time=t s
so, s=ut+0.5at^2
which would be, s= 4t+(0.5x0xt^2)
s=4t
And whereas, for Particle Y i took time= (t-2)s,
So, Particle Y: s= 6(t-2) + (0.5x(-2)x(t-2))
which is, s=10t-t^2-16
is that correct?
Last edited by Aleksander Krol; 1 month ago
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1 month ago
#3
Use SUVAT and write down the know quantities for both particle x and particle y. This will allow your to answer all of part a

(Original post by Aleksander Krol)
for part A: i did it as:
Ans: Particle X: u=4 m/s time=t s
so, s=ut+0.5at^2
which would be, s= 4t+(0.5x0xt^2)
s=4t
And whereas, for Particle Y i took time= (t-2)s,
So, Particle Y: s= 6(t-2) + (0.5x(-2)x(t-2))
which is, s=10t-t^2-16
is that correct?
Almost but there is one problem, it asks for the equations where t is the time after y has passed through O. Therefore the t for particle y is just t. And the t for particle x will have to be t+2
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#4
(Original post by tej3141)
Use SUVAT and write down the know quantities for both particle x and particle y. This will allow your to answer all of part a

Almost but there is one problem, it asks for the equations where t is the time after y has passed through O. Therefore the t for particle y is just t. And the t for particle x will have to be t+2
oh okay, thanks! but can you please explain how to do part b, please? if you can. i don't get ms's method.
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1 month ago
#5
(Original post by Aleksander Krol)
oh okay, thanks! but can you please explain how to do part b, please? if you can. i don't get ms's method.
Don't know what the markscheme has done.

If is the displacement of X, and similarly for Y, then the distance between them is , and that's what you want to minimise.

This can be a little tricky with a modulus sign there, so it's often the case that we square that and minimise the square (the minimum value will occur at the same value of t, although the function will give you the square of the distance between them.)

So, we minimise which is the same as , i.e. without the modulus signs.
Last edited by ghostwalker; 1 month ago
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1 month ago
#6
(Original post by Aleksander Krol)
so i get (t-1)+7>=0 so the minimum value is 7 so t=7 then, do we place the value of t in Sx - Sy to get the distance, right?
Also, why did we square them?
I suggest you post your full working, as t=7 doesn't follow from what you've put.

Your value of t would go into |Sx - Sy|, as that's the distance between them.

As to squaring, I explained that in the previous post.

I suggest you also post the markscheme.
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1 month ago
#7
FYI
https://pmt.physicsandmathstutor.com...ercise%201.pdf
page 10

You can use the fairly obvious fact that sx > sy so you simply find the minimmum of the quadratic
sx-sy
Though as ghostwalker says, you may have to be careful (square, absolute value, roots, ...) if you cant say something like that.

Also note that the minimum distance occurs when they have the same velocity, so this must occur at t=1 as they both have 4m/s.
Last edited by mqb2766; 1 month ago
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#8
(Original post by mqb2766)
FYI
https://pmt.physicsandmathstutor.com...ercise%201.pdf
page 10

You can use the fairly obvious fact that sx > sy so you simply find the minimmum of the quadratic
sx-sy
Though as ghostwalker says, you may have to be careful (square it, or absolute value) if you cant say something like that.

Also note that the minimum distance occurs when they have the same velocity, so this must occur at t=1 as they both have 4m/s.
i know i'm kind of annoying, but please help me understand it. so this is what i've understood and please correct me if i'm wrong. first of all, we squared sx-sy because we need to find the shortest distance right? i used completing the square method and got d=(t-1)^2+7 And 7 is the minimum value because the coefficient of t^2 is positive so the quadratic curve U well, i'm not sure but is that how d= 7m, because it's the minimum value and they have asked us to find the shortest distance?
Last edited by Aleksander Krol; 1 month ago
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1 month ago
#9
(Original post by Aleksander Krol)
i know i'm kind of annoying, but please help me understand it. so this is what i've understood and please correct me if i'm wrong. first of all, we squared sx-sy because we need to find the shortest distance right? i used completing the square method and got d=(t-1)^2+7 And 7 is the minimum value because the coefficient of t^2 is positive so the quadratic curve U well, i'm not sure but is that how d= 7m, because it's the minimum value and they have asked us to find the shortest distance?
There is no need to square here. Simply use sx-sy as this is > 0 for all t Its a quadratic because sy is quadratic and sx is linear. sx-sy is the distance between x and y, for all t.
Then when you complete the square of the quadratic sx-sy you get the minimum point (1,7). So the minimum value of sx-sy is 7 and this occurs at t=1.

Here it is in desmos with the vertical line representing sx-sy which is the distance between the points
https://www.desmos.com/calculator/dyaymdjlfa
Last edited by mqb2766; 1 month ago
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#10
(Original post by mqb2766)
There is no need to square here. Simply use sx-sy as this is > 0 for all t Its a quadratic because sy is quadratic and sx is linear. sx-sy is the distance between x and y, for all t.
Then when you complete the square of the quadratic sx-sy you get the minimum point (1,7). So the minimum value of sx-sy is 7 and this occurs at t=1.
thanks for bearing with me! <3<3<3
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#11
(Original post by ghostwalker)
I suggest you post your full working, as t=7 doesn't follow from what you've put.

Your value of t would go into |Sx - Sy|, as that's the distance between them.

As to squaring, I explained that in the previous post.

I suggest you also post the markscheme.
thanks for helping me! it's clear now <3 :,))
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1 month ago
#12
(Original post by Aleksander Krol)
thanks for bearing with me! <3<3<3
I added a desmos picture as you replied to the previous post
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#13
(Original post by mqb2766)
I added a desmos picture as you replied to the previous post
omg yeah, i've seen it now. i really appreciate your efforts in helping me understand it in the easiest way as possible. seriously thank you!
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1 month ago
#14
(Original post by Aleksander Krol)
omg yeah, i've seen it now. i really appreciate your efforts in helping me understand it in the easiest way as possible. seriously thank you!
In this case, the easiest way mathemetically to find the time of closest approach is when ]the velocities are the same so
6 - 2t = 4
This is because before that, y is getting closer to x (its velocity is larger) and after that y is getting further away from x (its velocity is smaller). So the time of closest approach is when the velocities are the same.

However, its "better" to do the sx-sy subtraction and complete the square, even though its a bit more maths. Part a) is a strong hint that you should do this as you model both sx and sy and it gives you both the time and distance of the closest approach directly from the completed square form.
Last edited by mqb2766; 1 month ago
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#15
(Original post by mqb2766)
In this case, the easiest way mathemetically to find the time of closest approach is when ]the velocities are the same so
6 - 2t = 4
This is because before that, y is getting closer to x (its velocity is larger) and after that y is getting further away from x (its velocity is smaller). So the time of closest approach is when the velocities are the same.

However, its "better" to do the sx-sy subtraction and complete the square, even though its a bit more maths. Part a) is a strong hint that you should do this as you model both sx and sy and it gives you both the time and distance of the closest approach directly from the completed square form.
sure, noted!
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