`s=f(t)=490t^2`

Here is the graph of s (displacement) against time (in seconds) . We see that the velocity (equivalent to the slope of the tangent of the curve) is not constant. At the beginning, the slope is `0` (the curve is horizontal, or momentarily flat). As time goes on, the object speeds up, and the slope of the curve gets steeper (more vertical).

Graph of displacement against time, giving velocity

Now the velocity is given by:

`v=lim_(h->0)(f(t+h)-f(t))/h`

So we have:

`v=lim_(h->0)(f(t+h)-f(t))/h`

`=lim_(h->0)([490(t+h)^2]-[490t^2])/h`

`=lim_(h->0)([490(t^2+2ht+h^2)]-[490t^2])/h`

`=lim_(h->0)([490(2ht+h^2)])/h`

`=lim_(h->0)[490{2t+h}]`

`=980t`

So `v = 980t` is the expression that tells us what the velocity is at any time (`t ≥ 0`).

When `t= 10` s, `v = 980(10) = 9800\ "cm/s"`.

So the velocity at `t= 10` s is `98\ "m/s"`.

We write velocity as: `v=(ds)/dt` OR we can also write this as: `v=s`'.

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