Question:
A cyclist starts cycling at \(t=0\) from \(x=0\). For the first ten seconds, her journey can be described by the equation \[x=4\sqrt{5t} + 6t^2\]
i) Find the cyclist's velocity after 5 seconds.

ii) Find her acceleration after 5 seconds.

Answer:

i)

The cyclist's velocity is \(\frac{dx}{dt}\), which can be found by differentiating the expression for \(x\) found above.

First, change the surd to an exponent.
\[x= 4\sqrt{5} t^\frac{1}{2} + 6t^2\] Now differentiate x with respect to t to find the general equation for \(\frac{dx}{dt}\).

\[\frac{dx}{dt}=4\sqrt {5} (\frac{1}{2})t^{-\frac{1}{2}} + (6)(2)t = 2\sqrt {\frac{5}{t}} + 12t\] The velocity after five seconds is found by subbing 5 in for \(t\) in the above expression.
\[\frac{dx}{dt}(5)= 2\sqrt {\frac{5}{5}} + 12(5)= 2+60 = 62ms^{-1}\] Make sure to include the units!

ii)

The acceleration can be found by differentiating the general expression for \(\frac{dx}{dt}\).
If \[\frac{dx}{dt}=2\sqrt {5}t^{-\frac{1}{2}} + 12t \] then
\[\frac{d^2x}{dt^2}= 2\sqrt {5} (-\frac{1}{2})t^{-\frac{3}{2}} + 12 = -\sqrt {\frac{5}{t^3}} + 12\] The acceleration after 5 seconds can be found by subbing 5 into the general expression: \[\frac{d^2x}{dt^2}= -\sqrt {\frac{5}{5^3}} + 12=-\sqrt {\frac{5}{125}} + 12=- {\frac{1}{5}} + 12 = \frac{59}{5}ms^{-2}\]

Second Order Derivatives