From this question, i understand that i must integrate "v" to find the displacement equation. But i thought that when you integrate, there had to be a "+C" in the equation. In the answer i was given, "C" was equal to zero. Can someone please justify this. thanks.

Question:
The velocity, after t seconds, of a particle is given in metres per second by
v = 3t^2 +1. Find distance travelled by the particle in the first 4 seconds.

Thank you.

2. Originally Posted by chrisgo
From this question, i understand that i must integrate "v" to find the displacement equation. But i thought that when you integrate, there had to be a "+C" in the equation. In the answer i was given, "C" was equal to zero. Can someone please justify this. thanks.

Question:
The velocity, after t seconds, of a particle is given in metres per second by
v = 3t^2 +1. Find distance travelled by the particle in the first 4 seconds.

Thank you.
velocity is the derivative of distance, so to get distance, we integrate velocity. We want to go from zero seconds to 4 seconds, so we set those values:

$\displaystyle \int_0^4 (3t^2+1) ~dt$

$\displaystyle =[t^3+t+C]_0^4$

$\displaystyle =[4^3+4+C]-[0^4+0+C]$

$\displaystyle =4^3+4+C-C$

$\displaystyle =68 ~metres$

You can see that the value of the constant is not relevant, because it will always be added and then subtracted. So when calculating definite integrals, we usually ignore it.

3. Originally Posted by angel.white
velocity is the derivative of distance, Mr F says: No, not in general. Only true when the direction of the velocity does not change. In general, velocity is the derivative of displacement (with respect to time).

so to get distance, we integrate velocity. We want to go from zero seconds to 4 seconds, so we set those values:

$\displaystyle \int_0^4 (3t^2+1) ~dt$

$\displaystyle =[t^3+t+C]_0^4$

$\displaystyle =[4^3+4+C]-[0^4+0+C]$

$\displaystyle =4^3+4+C-C$

$\displaystyle =68 ~metres$

You can see that the value of the constant is not relevant, because it will always be added and then subtracted. So when calculating definite integrals, we usually ignore it.
NB: This solution only works because the direction of velocity does not change. It wouldn't work if, for example, $\displaystyle v = 3t^2 - 1$.

If $\displaystyle v = 3t^2 - 1$ the best approach would be to consider the magnitude of the area between the curve $\displaystyle v = 3t^2 - 1$ and the t-axis between $\displaystyle 0 \leq t \leq 4$.