You are not logged in.
#1 2005-11-17 12:24:17
- CloudV3
- Novice
Offline
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 (2005-11-17 12:56:40)
#2 2005-11-17 18:26:51
- ganesh
- Moderator
Offline
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 
Character is who you are when no one is looking.
#3 2005-11-25 02:50:16
- 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.