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

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,930

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

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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**irspow****Guest**

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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