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#1 2005-11-16 13:24:17

CloudV3
Member
Registered: 2005-11-16
Posts: 1

Yet another integration problem...

Hi all, im new here and desparately need some help, itl probably be tooe asy for you guys, but i just cant seem to find the solution, so here goes...

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.


...Now, integrating v to get 5/3t^3 +t^2 + C.  Now how would i find out what C is equal to? Any help would be greatly appreciated!

Last edited by CloudV3 (2005-11-16 13:56:40)

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#2 2005-11-16 19:26:51

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,534

Re: Yet another integration problem...

v = 5t²+2t
vdt = (5t²+2t)dt
Integrating,
vt = 5t³/3  + t² + c
vt = S,
S = 5t³/3  + t² + c,
When time is zero, displacement is zero.
Therefore, 0 = 0 + 0 + c
c=0
S = 5t³/3  + t²
Put t = 2,
S = 40/3 + 4 = 52/3
Put t=1,
S= 5 /3+ 1 = 8/3
Therefore, the distance travelled in the 2nd second = 52/3-8/3 = 44/3 smile


Character is who you are when no one is looking.

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#3 2005-11-24 03:50:16

irspow
Guest

Re: Yet another integration problem...

Actually the constant C is not necessarily 0 in this case.  True, velocity is 0 at t=0, but the C of integration of the velocity function represents the inital position of the particle at t=0.  This was not specifically stated in the question.
     
     I will grant you that it does not matter in this question because any value of C would produce the same answer.  I just didn't want the incorrect explanation of C to go uncorrected.  Again, upon integration of a velocity function the value C will always be the initial position not velocity at t=0.

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