You are moving along a straight line, with your acceleration given by \[a(t) = 6t - 2.\]

- How fast will you be travelling at \(t=2\)?

You can think about the relationship between acceleration, velocity and displacement in two ways.

\[a(t) = \dfrac{d}{dt}\big(v(t)\big)\]

\[v(t) = \dfrac{d}{dt}\big(s(t)\big)\]

\[s(t) = \int{v(t)}\, dt\]

\[v(t) = \int{a(t)}\,dt\]

So we need to integrate the acceleration function to find your velocity function.

\[\begin{align*} v(t) &= \int{(6t - 2)}\,dt\\ v(t) &= 3t^2 - 2t + c \end{align*}\]What does the constant \(c\) represent in terms of velocity?

We know that \(t=2\). Do we have enough information to say how fast we are going?

If we calculate \(v(2)\) we get

\[\begin{align*} v(2) &= 12 - 4 + c\\ &= 8+c, \end{align*}\]so we could say our velocity is \(8 + c\) units.

Or we could calculate the integral

\[\begin{align*} \int_{0}^{2}{(6t - 2)}\,dt &= \left[3t^2 - 2t +c \right]^2_0\\ &= v(2) - v(0) \\ &= 8 \end{align*}\]and say our velocity is \(8\) units faster than it was at \(t = 0\).

In both cases we have accounted for the constant of integration which represents our initial velocity. This is important as we weren’t given any information about our velocity at \(t=0\), so we cannot assume that it was zero.

- Where will you be on the line at \(t=2\)?

To then find your displacement function we need to integrate the velocity function.

\[\begin{align*} s(t) &= \int{(3t^2 - 2t + c)}\,dt\\ s(t) &= t^3 - t^2 + ct + d \end{align*}\]Why is it important not to forget the constant of integration from the velocity function?

What does the constant \(d\) represent in terms of displacement?

With the displacement function we could calculate

\[\begin{align*} s(2) &= 2^{3} - t^{2} + 2c + d \\ &=4 + 2c +d \end{align*}\]or

\[\begin{align*} \int_{0}^{2}{(3t^2 - 2t + c)}\,dt &= \left[t^3 - t^2 + ct + d \right]^2_0 \\ &= s(2) - s(0) \\ &= 4 + 2c. \end{align*}\]One of these gives us your displacement from your starting point, and one gives us your displacement relative to a fixed origin. Which is which?

Under what circumstances would the two answers be the same?