• Dec 3rd 2006, 06:53 AM
c00ky
Hi guys,

I have the following problem:

At time = 0, a body is projected from an origin O with an initial velocity of 10ms-¹. The body moves along a straight line with a constant acceleration of -2ms-².

a) Find the displacement of the body from O when t = 7.
I have done this OK

b) How far from O does the body come to a instantaneous rest and what is the value of t then?
I have done this OK

c) Find the distance travelled by the body during the time interval t = 0 to t = 7.
I do not know this, help please
• Dec 3rd 2006, 07:02 AM
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,
$\displaystyle -t^2+10t |_0^5 +10t-t^2|_5^7$
• Dec 3rd 2006, 10:50 AM
topsquark
Quote:

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

Everything ThePerfectHacker said was correct,

EXCEPT

he's got the roles of distance and displacement reversed.
$\displaystyle \int_0^7 -2t+10dt$ <--Gives the displacement.

$\displaystyle \int_0^7 |-2t+10| dt$<--Gives the distance.

-Dan