Originally Posted by

**ThePerfectHacker** Where are told,

$\displaystyle a(t)=-2\frac{\mbox{m}}{\mbox{sec}^2}$

The the velocity is the integral of that,

$\displaystyle v(t)=-2t+C$

But the initial conditions say,

$\displaystyle v(0)=10$

Thus,

$\displaystyle v(t)=-2t+10$

So the distance is,

$\displaystyle \int_0^7 -2t+10dt=-t^2+10t|_0^7=-49+70=21$

The displacement is,

$\displaystyle \int_0^7 |-2t+10| dt$

This integral is more confusing to integrate.

But the trick is to divide it into parts.

One where the $\displaystyle -2t+10>0$ and the other when $\displaystyle -2t+10<0$

When does this change happen?

Simple find,

$\displaystyle -2t+10<0$

$\displaystyle -2t<-10$

$\displaystyle t>5$

So at $\displaystyle t=5$ we have a change.

Thus,

$\displaystyle \int_0^7 |-2t+10|dt=\int_0^5 |-2t+10|dt+\int_5^7 |-2t+10|dt$

But because of the changing signs we can write,

$\displaystyle \int_0^5 -2t+10dt+\int_5^7 10-2t dt$

Thus,

$\displaystyle -t^2+10t |_0^5 +10t-t^2|_5^7 $