S U V A T, please help.

**c00ky** · December 3rd 2006, 06:53 AM

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

**ThePerfectHacker** · December 3rd 2006, 07:02 AM

Where are told,
$a(t)=-2\frac{\mbox{m}}{\mbox{sec}^2}$
The the velocity is the integral of that,
$v(t)=-2t+C$
But the initial conditions say,
$v(0)=10$
Thus,
$v(t)=-2t+10$

So the distance is,
$\int_0^7 -2t+10dt=-t^2+10t|_0^7=-49+70=21$

The displacement is,
$\int_0^7 |-2t+10| dt$
This integral is more confusing to integrate.
But the trick is to divide it into parts.
One where the $-2t+10>0$ and the other when $-2t+10<0$
When does this change happen?
Simple find,
$-2t+10<0$
$-2t<-10$
$t>5$
So at $t=5$ we have a change.
Thus,
$\int_0^7 |-2t+10|dt=\int_0^5 |-2t+10|dt+\int_5^7 |-2t+10|dt$
But because of the changing signs we can write,
$\int_0^5 -2t+10dt+\int_5^7 10-2t dt$
Thus,
$-t^2+10t |_0^5 +10t-t^2|_5^7$

**topsquark** · December 3rd 2006, 10:50 AM

Originally Posted by ThePerfectHacker

Where are told,
$a(t)=-2\frac{\mbox{m}}{\mbox{sec}^2}$
The the velocity is the integral of that,
$v(t)=-2t+C$
But the initial conditions say,
$v(0)=10$
Thus,
$v(t)=-2t+10$

So the distance is,
$\int_0^7 -2t+10dt=-t^2+10t|_0^7=-49+70=21$

The displacement is,
$\int_0^7 |-2t+10| dt$
This integral is more confusing to integrate.
But the trick is to divide it into parts.
One where the $-2t+10>0$ and the other when $-2t+10<0$
When does this change happen?
Simple find,
$-2t+10<0$
$-2t<-10$
$t>5$
So at $t=5$ we have a change.
Thus,
$\int_0^7 |-2t+10|dt=\int_0^5 |-2t+10|dt+\int_5^7 |-2t+10|dt$
But because of the changing signs we can write,
$\int_0^5 -2t+10dt+\int_5^7 10-2t dt$
Thus,
$-t^2+10t |_0^5 +10t-t^2|_5^7$

Everything ThePerfectHacker said was correct,

EXCEPT

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

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

-Dan

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S U V A T, please help.