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# M2 Quick Question watch

1. How do you do this Q:

A particle moves in a straight line with velocity v given by at time t s after passing throught a fixed point O in that line.

I know that the velocity is 0 when t=1

and the accerelation is -45 when t=2

Find

c) maximum velocity ANS=2

d) its distance from O when t=2 ANS=-8

On C, don't you differientate v to get a, make it equal to 0, then solve, you should get 2 but I get a 1/2

Don't know what to do a d)
2. (c) asks for maximum velocity. You've found what t is, now work out v
(d) you need to integrate the velocity to find the displacement
3. (Original post by emmaxoxo)
How do you do this Q:

A particle moves in a straight line with velocity v given by at time t s after passing throught a fixed point O in that line.

I know that the velocity is 0 when t=1

and the accerelation is -45 when t=2

Find

c) maximum velocity ANS=2

d) its distance from O when t=2 ANS=-8

On C, don't you differientate v to get a, make it equal to 0, then solve, you should get 2 but I get a 1/2

Don't know what to do a d)
for c), yes you do get a half, but this is the time. So just sub it back into the velocity formula to get the velocity.

For D, integrating V gives X (when dealing with t)
4. oh right thanks got it
5. (Original post by ian.slater)
(c) asks for maximum velocity. You've found what t is, now work out v
(d) you need to integrate the velocity to find the displacement
Ok, New Question

Particle P moves in a striaght line such that t s its accerleation is given by a=2t-2, it passes through a point O on the line t=0 with v=1

find the distance of P from O when it's at rest.

Do you find v, integrate it to get x, then sub t=0, to find x at rest??? I get the answer 0 whereas it should be a 1/3
6. (Original post by emmaxoxo)
Ok, New Question

Particle P moves in a striaght line such that t s its accerleation is given by a=2t-2, it passes through a point O on the line t=0 with v=1

find the distance of P from O when it's at rest.

Do you find v, integrate it to get x, then sub t=0, to find x at rest??? I get the answer 0 whereas it should be a 1/3
No, you integrate a to find V, then you equate that to 0, since you want V to be 0. Ie when it is at rest v=0. This will give u the value of t.

Integrate V again to give X formula, then sub the value of t into the x formula to get the Distance
7. (Original post by 2710)
No, you integrate a to find V, then you equate that to 0, since you want V to be 0. Ie when it is at rest v=0. This will give u the value of t.

Integrate V again to give X formula, then sub the value of t into the x formula to get the Distance
OP - Don't forget when you integrate a to get v, that you need to find out what the constant of integration is. You're told that v=1 when t=0 so use that.
8. (Original post by 2710)
No, you integrate a to find V, then you equate that to 0, since you want V to be 0. Ie when it is at rest v=0. This will give u the value of t.

Integrate V again to give X formula, then sub the value of t into the x formula to get the Distance
Thanks. Got it.

Last Question.

A particle moves along the x axis with velocity at time t given by v=5t^2+2t

find the distance it moves in the 2nd second (i.e. t=2)

Don't you just intergrate v to find x. But how do you find out what C is on x, what are the restraints (i.e. x=0,t=0), therefore c=0, then just sub t=2 into x. I get 17 and 1/3, whereas it should be 14 2/3?
9. (Original post by emmaxoxo)
Thanks. Got it.

Last Question.

A particle moves along the x axis with velocity at time t given by v=5t^2+2t

find the distance it moves in the 2nd second (i.e. t=2)

Don't you just intergrate v to find x. But how do you find out what C is on x, what are the restraints (i.e. x=0,t=0), therefore c=0, then just sub t=2 into x. I get 17 and 1/3, whereas it should be 14 2/3?
So you have to finds the distance between times 2-3 s. So x=0 when t=2. Find c with that, and try it again. I think this is how you do it anyways.
10. Distance moved in 2nd second is x(2) - x(1)

Fortunately, whatever constant of integration you use will disappear when you work this out. In other words you can use whatever original you want.

If I were doing this question, I'd be nit-picky enough to check that the particle doesn't change direction between t=1 and t=2, but I don't expect M1 is that awkward!
11. (Original post by 2710)
So you have to finds the distance between times 2-3 s. So x=0 when t=2. Find c with that, and try it again. I think this is how you do it anyways.
what so find the distance it moves in the 2nd second does not mean t=2, it means the distance between second 2 and 3 ???

So C= - 17 and 1/3

then you sub in t=3 in X and t=2 in X again, and subtract them? if so i'm still getting the wrong ANS
12. (Original post by ian.slater)
Distance moved in 2nd second is x(2) - x(1)

Fortunately, whatever constant of integration you use will disappear when you work this out. In other words you can use whatever original you want.

If I were doing this question, I'd be nit-picky enough to check that the particle doesn't change direction between t=1 and t=2, but I don't expect M1 is that awkward!
Sorry I keep getting forgetting this is M2.

Ok I have got the amswer right by saying the C=0, this true then? But why does C=0 ??
13. Woops sorry, yeh, ian is right. Between the 1 and 2nd second
14. (Original post by emmaxoxo)
Sorry I keep getting forgetting this is M2.

Ok I have got the amswer right by saying the C=0, this true then? But why does C=0 ??
Because the question doesn't tell you where the particle is at t=0, or enough information to work it out, you can be sure that they will not ask you anything that needs you to know that.

So you can say 'let x=0 when t=0' i.e. set your own origin for measurement.

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