Mechanics help

KallamSamad

I’ll show attachments.
But if you integrate and find c and integrate again is c=0?

Reply 1

KallamSamad

Why is the answer 24m when I got 240m

Reply 2

mosaurlodon

Well, if you got part a, it should integrate into the answer in part c - maybe check your signs?

Im pretty sure the reason they ignore +c is because if you take the integral from 0-6 you end up cancelling the +c, since you get (24+c) - (0+c) = 24

Reply 3

chavvo

Original post by KallamSamad

I’ll show attachments.
But if you integrate and find c and integrate again is c=0?

Not entirely sure what you're asking, but as the previous poster hinted, the constant of integration only applies for an indefinite integral. If you're evaluating a definite integral then in effect you can ignore the constant because it cancels out in the working.

Reply 4

mqb2766

Original post by KallamSamad

I’ll show attachments.
But if you integrate and find c and integrate again is c=0?

I think the first integration youre talking about is from acceleration to velocity in which case the initial condition is
v(0) = 10
Which is the constant of integration (c=10) for this integration problem.

The second integration youre talking about(?) is from velocity to displacement and the initial condition is
s(0) = 0
as the particle passes through the origin at time 0. So the constant of integration for this problem is c=0.

They are different integration problems with seperate constants of integration.