Definite intergrals for an expression of displacement?

**jonoliv8** · October 20th 2012, 12:13 AM

Hi I'm just getting back into education and I've been given some practice questions but this one kind on thrown me.
Ive been given the expression
v=12te^(t/2) +5 for velocity

I had work out what the velocity is at t= 4.5 s
i got 10.69m/s

Then I had to obtain an expression for acceleration and calculate the force acting on a body of 5kg at t = 4.5s so I used product rule and finally got an answer.

but the last part of the question again asks for an expression this time for displacement s of the body after t seconds, given that when t=0, s=0. Hence calculate the displacement of the body at time t=4.5 seconds
I was thinking definite intergrals but its been a while since I studied it so any ideas please?

**MarkFL** · October 20th 2012, 12:36 AM

Given that you report finding $v(4.5)\approx10.69\,\frac{\text{m}}{\text{s}}$ I assume we have:

$v(t)=12te^{-\frac{t}{2}}+5$

Since velocity is defined as the time rate of change of displacement, we may write the initial value problem:

$\frac{ds}{dt}=12te^{-\frac{t}{2}}+5$ where $s(0)=0$

and so integrating with respect to $t$ , using the boundaries as limits, and using dummy variables of integration, we may state:

$\int_0^{s(t)}\,du=\int_0^t 12ve^{-\frac{v}{2}}+5\,dv$

For the first term in the integrand on the right, you will want to use integration by parts. Are you familiar with this method?

**jonoliv8** · October 20th 2012, 01:05 AM

I've not touched on any integration for a few years just reading up on it now thanks for the reply

**jonoliv8** · October 20th 2012, 04:15 PM

Ok I've tried to use integration by parts but keep going down dead ends this is where I'm at:
I= 12t^-0.5t
I= 12t x e^-0.5t/-0.5 - { e^-0.5/-0.5 x 12dt
I= 12te^-0.5t/-0.5. - 12/-0.5 x e^-0.5t/-0.5
I= 12te-0.5t/-0.5 - 12e^-0.5t/0.25
Is this correct before I definate integrate and do I add the original +5 back in?

**MarkFL** · October 20th 2012, 06:08 PM

I would let:

$I_1=12\int_0^t we^{-\frac{w}{2}}\,dw$

$I_2=5\int_0^t\,dv=5t$ (this one we are able to evaluate directly).

So, we are left with the first term on which to use integration by parts.

Let:

$u=w\,\therefore\,du=dw$

$dv=e^{-\frac{w}{2}}\,dw\,\therefore\,v=-2e^{-\frac{w}{2}}$

and so we have:

$\frac{I_1}{12}=\left(\left[-2we^{-\frac{w}{2}} \right]_0^t+2\int_0^t e^{-\frac{w}{2}}\,dw$

$\frac{I_1}{12}=-2te^{-\frac{t}{2}}+2\left[-2e^{-\frac{w}{2}} \right]_0^t$

$\frac{I_1}{12}=-2te^{-\frac{t}{2}}+2\left(-2e^{-\frac{t}{2}}+2 \right)$

$\frac{I_1}{12}=-2te^{-\frac{t}{2}}-4e^{-\frac{t}{2}}+4$

$I_1=48-24e^{-\frac{t}{2}}(t+2)$

Add so, putting everything together, we find:

$s(t)=5t+48-24e^{-\frac{t}{2}}(t+2)$

**jonoliv8** · October 21st 2012, 04:28 AM

Ok but when t=0 s= 48 and it should equal zero also that's what confused me with my answer?

**jonoliv8** · October 21st 2012, 04:52 AM

Ignore that, finger trouble

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