
Kinematics problem
A remote control car starts moving in a straight line from a fixed point. The velocity in metres per second is given by the formula
v = 3t^{2}  4t  8
i) what is the initial acceleration?
So for this question I assume you differentiate, to get your acceleration equation: a = 6t  4
Then I'd assume you put the t at 0, to get the initial acceleration, because that is what initial means  at time 0. But that just gives me 4 acceleration, which doesn't seem right to me.
ii) how far does the car travel in the third second?
I'd assume you integrate, on this one, because you want the distance formula: d = t^{3}  2t^{2}  8t + c
The answer I got from this was that the constant was 15, because I set the letter t to 3, and, yup.
But then, the distance turned out to equal, um, 0. Which also didn't seem right.
What am I doing wrong, in both questions, that I end up with these weird answers?? (Thinking)

Re: Kinematics problem
For the initial acceleration, the negative sign just indicates the direction of the acceleration, which depends on how you have defined your coordinate axis.
For the distance traveled in the 3rd second, you want to evaluate:
$\displaystyle \int_2^3v(t)\,dt$
You should note that the velocity changes sign in this interval, so you want to find the root of the velocity function in this interval, and using the definition of absolute value, break this up into two integrals. Without the absolute value, what you are finding is the displacement, rather than the distance traveled, which is 1 meter.