A body moves x metres in t seconds, where $\displaystyle x = t^3-7t^2+3
$
Find:
{i} the velocity at t = 6 seconds
{ii} the acceleration at t = 8 seconds.
Hello,
1. I assume that you are modeling a continuously(?) accelerated motion. (I'm pretty sure that this expresion is not accurate but I don't have a special (Physics) dictionary)
2. The value of x is the distance after t seconds of movement:
$\displaystyle d(t) = t^3-7t^2+3$
3. The speed after t seconds is:
$\displaystyle v(t) = d'(t) = 3t^2-14t~\Longrightarrow~ v(6) = 24\frac{m}{s}$
4. The acceleration is:
$\displaystyle a(t) = v'(t) = d''(t) = 6t-14~\Longrightarrow~ a(8) = 34\frac{m}{s^2}$
absolutely right, just remember that
Displacement ----> Velocity(Displacement/Time) ----> Acceleration (Change of Speed/Time) , Where the arrows indicates a derivative (e.g. the derivative of Displacement gives Velocity)
If you need to go the other way, just integrate, remembering to add c, which you will have to work out from the info given by the question