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?