# The position of an object moving along a line is given by p(t) = t^2 - 6t +3. What is the speed of the object at t = 3 ?

Mar 5, 2017

As the speed is the derivative of the position function, at $t = 3$, its speed is zero.

#### Explanation:

I have to assume you are working with calculus in this Physics course. The first derivative of the position with respect to time will give the velocity function:

$\frac{\mathrm{dp} \left(t\right)}{\mathrm{dt}} = v \left(t\right)$

$\frac{d}{\mathrm{dt}} \left({t}^{2} - 6 t + 3\right) = 2 t - 6$

If we evaluate this function at $t = 3$

$v = 2 \left(3\right) - 6 = 0$

The object has (momentarily) stopped at $t = 3$ s. (But note that this does not imply the position or the acceleration is zero. In fact it's position is ${3}^{2} - 6 \left(3\right) + 3 = - 6$

And, for the record, its acceleration is the second derivative of $p \left(t\right)$ or the first derivative of $v \left(t\right)$, namely 2.