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Whateverisbest
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A particle P of mass 0.2kg is at rest at a fixed origin O. At time t seconds where 0≦t≦3, a force of (2ti + 3j)N is applied to P.
a) find the position vector of P when t=3
b) When t=3, the force acting on P changes to (6i + (12 - t^2)j)N, where 3
≦t.

I just can't get the same answer as the back of the book for b).

After working out the acceleration, I integrated to get the velocity function:
30ti + (60t - (5t^3)/3 )j +C

To work out C I used the acceleration in a) to get a velocity -> 5t^2 i +15tj.

Then by making t=0 and v= 45i + 45j, I got C=45i + 45j.

Making t=3 in v= (30t+45)i + (60t +45 -(5t^3)/3)j , I get v= (135i + 180j)
The answer on the back states (135i - 90j) which I'm unable to get.

Could any one point me in the right direction? I'll post my working on paper if the above doesn't make sense.

Thanks in advance.
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RoyalBlue7
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(Original post by Whateverisbest)
A particle P of mass 0.2kg is at rest at a fixed origin O. At time t seconds where 0≦t≦3, a force of (2ti + 3j)N is applied to P.
a) find the position vector of P when t=3
b) When t=3, the force acting on P changes to (6i + (12 - t^2)j)N, where 3
≦t.

I just can't get the same answer as the back of the book for b).

After working out the acceleration, I integrated to get the velocity function:
30ti + (60t - (5t^3)/3 )j +C

To work out C I used the acceleration in a) to get a velocity -> 5t^2 i +15tj.

Then by making t=0 and v= 45i + 45j, I got C=45i + 45j.

Making t=3 in v= (30t+45)i + (60t +45 -(5t^3)/3)j , I get v= (135i + 180j)
The answer on the back states (135i - 90j) which I'm unable to get.

Could any one point me in the right direction? I'll post my working on paper if the above doesn't make sense.

Thanks in advance.
Firstly I assume they are asking to find the velocity when t=3, but there's a problem with this. You haven't posted the entire question!

Secondly when you use t=0 when calculating for C the velocity has to be 0 and not 45i + 45j.

For part b) are they asking to find the velocity at t=3? How did you get 45i + 45 j?
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elMaestro_8
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(Original post by Whateverisbest)
A particle P of mass 0.2kg is at rest at a fixed origin O. At time t seconds where 0≦t≦3, a force of (2ti + 3j)N is applied to P.
a) find the position vector of P when t=3
b) When t=3, the force acting on P changes to (6i + (12 - t^2)j)N, where 3
≦t.

I just can't get the same answer as the back of the book for b).

After working out the acceleration, I integrated to get the velocity function:
30ti + (60t - (5t^3)/3 )j +C

To work out C I used the acceleration in a) to get a velocity -> 5t^2 i +15tj.

Then by making t=0 and v= 45i + 45j, I got C=45i + 45j.

Making t=3 in v= (30t+45)i + (60t +45 -(5t^3)/3)j , I get v= (135i + 180j)
The answer on the back states (135i - 90j) which I'm unable to get.

Could any one point me in the right direction? I'll post my working on paper if the above doesn't make sense.

Thanks in advance.

I used v=u+at in part B and the answer I got was (135i + 90j)

the difference between my answer and the one in the back of your book is the sign between the I and j vectors where I got a + sign and it must be a - sign.

so can you double check the answer in your book, and in case of my answer being correct, I used v=u+at ,where

v: velocity after new force
u: velocity before new force (5t^2i+15tj)
a: acceleration after the mew force [30i+(60-5t^2)j]
t: time (t)

by substituting and simplifying I get v=(5t^2+30t)I+(75t-5t^3)j

by taking t=3

v=135i +90j


but in case my answer isn't correct, please post your working in a paper and show me the question and answer to be able to help you
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elMaestro_8
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(Original post by Whateverisbest)
A particle P of mass 0.2kg is at rest at a fixed origin O. At time t seconds where 0≦t≦3, a force of (2ti + 3j)N is applied to P.
a) find the position vector of P when t=3
b) When t=3, the force acting on P changes to (6i + (12 - t^2)j)N, where 3
≦t.

I just can't get the same answer as the back of the book for b).

After working out the acceleration, I integrated to get the velocity function:
30ti + (60t - (5t^3)/3 )j +C

To work out C I used the acceleration in a) to get a velocity -> 5t^2 i +15tj.

Then by making t=0 and v= 45i + 45j, I got C=45i + 45j.

Making t=3 in v= (30t+45)i + (60t +45 -(5t^3)/3)j , I get v= (135i + 180j)
The answer on the back states (135i - 90j) which I'm unable to get.

Could any one point me in the right direction? I'll post my working on paper if the above doesn't make sense.

Thanks in advance.
Hello again

I found that question in my book while I was revising. My book was similar to yours after all.
while I was solving the question I remembered your thread, and came back to help you since I got the correct answer.

I already colored your mistakes by red above, but to make it clearer for you I'd explain your mistakes for you.

First of all you have taken t=0 when v=45i+45j ,while you were expected to take t=3, because v is equal to 45i+45j when t is equal to 3. and if you worked that out you'll find c=(-45i-90j) .

Secondly, you mistakenly found v when t=3, while you should have taken t=6, because this is what the question asks for, and if you worked that out, you'll find that v=135i-90j


I hope that was helpful
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the bear
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since the force is not constant ( and the mass is constant ) the acceleration is not constant ... so you cannot use SUVAT
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