a(t) for a 1D motion of a point is given by a(t)=3t^2+5t6 for t<20 and a(t)=2t^3t+1 for t>20. Find s when t=32.
The question essentially wants s(t) then plug in t=32 to find s, I think. First step must be to find v when t=20, so let's integrate:
v=t^3+(5/2)t^26t for t<20 (constant=0)
Evaluate this when t=20: v=8880 ms^(1).
Now v(t) for t>20. v(t)=(1/2)t^4(1/2)t^2+t+C
Here's my first stumbling block: how do we find the value of C now?
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BigDaddy
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 14052013 18:26

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 14052013 18:43
(Original post by BigDaddy)
a(t) for a 1D motion of a point is given by a(t)=3t^2+5t6 for t<20 and a(t)=2t^3t+1 for t>20. Find s when t=32.
The question essentially wants s(t) then plug in t=32 to find s, I think. First step must be to find v when t=20, so let's integrate:
v=t^3+(5/2)t^26t for t<20 (constant=0)
Evaluate this when t=20: v=8880 ms^(1).
Now v(t) for t>20. v(t)=(1/2)t^4(1/2)t^2+t+C
Here's my first stumbling block: how do we find the value of C now? 
Smaug123
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 14052013 18:53
(Original post by BigDaddy)
how do we find the value of C now?(Original post by joostan)
The constant of integration is the initial velocity.
If, for example, we had , we'd have , and then the arbitrary constant wouldn't be the initial velocity, but (the initial velocity)1. 
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 14052013 18:55
(Original post by Smaug123)
Although note that this isn't necessarily true in general  it so happens that the velocity here is a polynomial, so when t=0, every term cancels out so that the arbitrary constant is the only term left.
If, for example, we had , we'd have , and then the arbitrary constant wouldn't be the initial velocity, but (the initial velocity)1. 
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 14052013 19:06
(Original post by joostan)
The constant of integration is the initial velocity.
v(t)=(1/2)t^4(1/2)t^2+t+C
We know that at t=20, v=8880. so C=8880((1/2)*20^4(1/2)*20^2+20)=70940. So v(t)=(1/2)t^4(1/2)t^2+t70940.
Now integrate the v(t) for t<20.
s(t)=(1/4)t^4+(5/6)t^33t^2 (constant=0)
Plug in t=20 to find s(20)=45,466.67
Integrate the v(t) for t>20. s(t)=(1/10)t^5(1/6)t^3=(1/2)t^270940t+C
Calculate C using s(20)=45,466.67 and t=20 in that equation.
This is the function s(t) for t>20. Plug in t=32 and you have t. No need to add any previous constants or previous distances, you just plug in 32 (once you've evaluated C) and you get out the value of the displacement s at t=32.
Correct? 
BigDaddy
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 14052013 19:07
(Original post by Smaug123)
Although note that this isn't necessarily true in general  it so happens that the velocity here is a polynomial, so when t=0, every term cancels out so that the arbitrary constant is the only term left.
If, for example, we had , we'd have , and then the arbitrary constant wouldn't be the initial velocity, but (the initial velocity)1.
Could you check through to see if my working makes sense? (The post above) 
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 14052013 20:03
(Original post by BigDaddy)
Could you check through to see if my working makes sense? (The post above) 
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 14052013 20:20
(Original post by Smaug123)
I haven't checked your numbers, but it looks fine to me. On a vaguely related note, something to be aware of is that they might try and catch you out by asking about distance rather than displacement  if you integrate velocity you'll get displacement, which isn't helpful in a distance question. 
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 14052013 20:48
(Original post by BigDaddy)
How would I solve this question if they asked for distance instead of displacement? 
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 14052013 22:18
(Original post by Smaug123)
Then you'd have to look at the velocity and make sure that whenever the velocity was negative, you continued to add on the distance (rather than subtract, as you would implicitly be doing with displacement). That is, . 
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 14052013 22:29

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 14052013 23:28
(Original post by Smaug123)
A dummy variable that I just use for integration; it could just as well be or something. I would usually use , except that x already has a meaning (in this example, it's distance) so I can't use it again without being a bit sloppier than I would like.
Are you telling me I can't just integrate the modulus of v(t) with respect to t and get my answer? 
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 15052013 00:37
(Original post by BigDaddy)
Isn't it ?
Are you telling me I can't just integrate the modulus of v(t) with respect to t and get my answer?
When you integrate, you integrate with respect to some variable. In the process, that variable "disappears" if you substitute stuff in (that is, if you're doing a definite integral). For example: , but  and you see that the x has vanished. Hence, if you do a definite integral, you need to use a "dummy variable": , and that's how you get t in your answer. Make sense? It's too late at night for me to be coherent :P 
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 15052013 00:46
(Original post by Smaug123)
Urgh, I find that this is quite a hardtoexplain point (a fact about my incompetence, I know, rather than the fact itself).
When you integrate, you integrate with respect to some variable. In the process, that variable "disappears" if you substitute stuff in (that is, if you're doing a definite integral). For example: , but  and you see that the x has vanished. Hence, if you do a definite integral, you need to use a "dummy variable": , and that's how you get t in your answer. Make sense? It's too late at night for me to be coherent :P
Maybe it would be best if you could just solve the example in my OP for distance? No need to get any numbers, just say the process. 
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 15052013 00:51
(Original post by BigDaddy)
That's all very well, in fact it makes perfect sense  except that in your previous post you wrote , which seems to conflict with the fact that you're still writing when integrating f(x) here.
Maybe it would be best if you could just solve the example in my OP for distance? No need to get any numbers, just say the process.
In the example, you need to do  of course, we don't necessarily have an initial condition any more, because the initial condition was about displacement rather than distance. You would carry out this integral by working out where the integrand became negative, splitting up the range of integration at those points, and evaluating the integrals separately there. Then you can add them together at the end. (Sorry, I'm a bit too tired at the moment to do it properly…)Last edited by Smaug123; 15052013 at 00:51. Reason: Horrible oddness 
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 15052013 00:53
(Original post by Smaug123)
In which case I shall rewrite my previous post as  the name I use for the variable doesn't matter, as long as it's a) consistent, and b) not used outside the integral.
In the example, you need to do  of course, we don't necessarily have an initial condition any more, because the initial condition was about displacement rather than distance. You would carry out this integral by working out where the integrand became negative, splitting up the range of integration at those points, and evaluating the integrals separately there. Then you can add them together at the end. (Sorry, I'm a bit too tired at the moment to do it properly…) 
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 15052013 01:29
(Original post by BigDaddy)
a(t) for a 1D motion of a point is given by a(t)=3t^2+5t6 for t<20 and a(t)=2t^3t+1 for t>20. Find s when t=32.
The question essentially wants s(t) then plug in t=32 to find s, I think. First step must be to find v when t=20, so let's integrate:
v=t^3+(5/2)t^26t for t<20 (constant=0)
Evaluate this when t=20: v=8880 ms^(1).
Now v(t) for t>20. v(t)=(1/2)t^4(1/2)t^2+t+C
Here's my first stumbling block: how do we find the value of C now? 
Smaug123
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 15052013 09:26
(Original post by TheGrinningSkull)
Quick question, those a's highlighted, what do they stand for? 
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 15052013 12:48
(Original post by Smaug123)
Acceleration.Last edited by TheGrinningSkull; 15052013 at 12:50.
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