Integration, Acceleration, Velocity and time

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Sep 14th 2008, 07:25 PM
immunofort

Integration, Acceleration, Velocity and time

The question states

An object is moving in a straight line through a fixed point A.
When the object first passes through A it is travelling at 5cm per sec and the timing beings.
The acceleration of the object is given by

a= 6t-16
where acceleration is in cm/sec^2 at time t seconds after the object first passes through A.
Find the distance of the object from A the second time it comes to rest.

So i first integrated to get V=3t-16+c
since t=0 , V=5 then c=5
so V=3t-16+c

Then i integrated again to get s = t^3 -8t^2 + 5t + c
I thought at
time = 0, distance would equal 0 therefore making c=0 aswell, but when i look at the answer its telling me s=3, so i just assumed that to be right and decided to think about it later as i knew when working out t for when V=0 it wouldn't matter, so i got x= 1/3 and 5 which is correct according to the answer.

then it says in the answer s(5) = -50cm

but when i input 5 into the equation s = t^3 -8t^2 + 5t + 3 it gives me -47
thus bringing me to the 3 again,

Where does the 3 come from? or is the answer book wrong? am i over looking something?
Sep 14th 2008, 10:34 PM
mr fantastic

Quote:

Originally Posted by immunofort

The question states

An object is moving in a straight line through a fixed point A.
When the object first passes through A it is travelling at 5cm per sec and the timing beings.
The acceleration of the object is given by

a= 6t-16
where acceleration is in cm/sec^2 at time t seconds after the object first passes through A.
Find the distance of the object from A the second time it comes to rest.

So i first integrated to get V=3t-16+c
since t=0 , V=5 then c=5
so V=3t-16+c

Then i integrated again to get s = t^3 -8t^2 + 5t + c
I thought at
time = 0, distance would equal 0 therefore making c=0 aswell, but when i look at the answer its telling me s=3, so i just assumed that to be right and decided to think about it later as i knew when working out t for when V=0 it wouldn't matter, so i got x= 1/3 and 5 which is correct according to the answer.

then it says in the answer s(5) = -50cm

but when i input 5 into the equation s = t^3 -8t^2 + 5t + 3 it gives me -47
thus bringing me to the 3 again,

Where does the 3 come from? or is the answer book wrong? am i over looking something?

$x = t^3 - 8t^2 + 5t$ NOT $x = t^3 - 8t^2 + 5t + 3$ .

I don't know where the 3 is coming from but it should not be in your expression for x since x(0) = 0 NOT 3.
Sep 15th 2008, 01:03 AM
immunofort

thanks

Ok thanks for that, Just needed some confirmation from somebody who was a bit more experienced than me. I guess it must be fault on the exam sheet.