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A stone is dropped from a cliff. The distance it has fallen is s = 4.9t² metres (m) after t seconds…

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A stone is dropped from a cliff. The distance it has fallen is $s = 4.9t^2$ metres (m) after $t$ seconds (s). A camera measures the average speed of the stone between $t = 2$ s and $t = 2.1$ s. A second instrument shows the speed at exactly $t = 2$ s. Find the average speed over the short interval. Then find the speed at $t = 2$ s, which is the limit of the average speeds as the interval shrinks.

Need a reminder? The key ideaShow

The derivative is the limit of the secant slopes: $f'(a) = \displaystyle\lim_{h \to 0} \dfrac{f(a + h) - f(a)}{h}$ . Also $(x^n)' = nx^{n-1}$ , $(cf)' = cf'$ and $(f + g)' = f' + g'$ . The tangent at $a$ is $y - f(a) = f'(a)(x - a)$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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