# A confusing velocity word problem?

So the question is this:

The distance $S$ in km of a bike rider can be shown through this formula:

$$S(t)= (t-3)^3+27$$

The domain of the function is $t$ in hours: $[0,5]$

How would you go about finding the slowest and fastest velocity of this bike rider?

I could find the slowest speed, which is at time $t=3$, but the fastest is somewhat tricky. I would appreciate if you could shed some light upon the matter. Thanks.

• What's a "velocity word problem"? Mar 20, 2017 at 22:23
• @user395550 The tricky part is to find out, at what time or moment is the velocity at its hightest? Mar 20, 2017 at 22:24
• Remember that looking for points at which the derivative vanishes only gives you potential local maxima/minima. There could also be a global max/min at at one or both of the endpoints of the domain.
– amd
Mar 21, 2017 at 0:12

$s(t) = (t-3)^3+27$

$v(t) = \frac{ds}{dt} = 3(t-3)^2$

$a(t) = \frac{dv}{dt} = 6(t-3)$

To find the the stationairy points of $v$, we set $a$ = $0$.

$6(t-3) = 0 \implies t = 3$

$v(3) = 0$ which is clearly the minimum possible speed.

To find the maximum, we must consider how the function $v(t)$ behaves.

$v(t) = 3(t-3)^2$ which we can see means that $|v|$ is increasing for $t>3$ and for $t<3$. Therefore, we must check the minimum and maximum values of $t$ in the domain, and see which one gives the larger value of $v$ - i.e. our maximum $v$.

$$|v(5)| = 12km/h$$

$$|v(0)| = 27km/h$$

Hence $v_{max} =27 km/h$

The trick is to take the second derivative: $S''(t)=(3(t-3)^2)'=6(t-3)$. As you said, the minimum of the speed corresponds to the zero of the second derivative (from another point of view, the function $S$ changes convexity in $x=3$). There aren't other zeros of the second derivative: so the maximum has to be searched on the extremes of the interval: $S'(0)=3*(0-3)^2=27,$ and $S'(5)=3*(5-3)^2=12$. So the maximum speed is the initial speed.

You seem to have figured out that the slowest velocity of the rider happens at time $t = 3$ Note that the equation for the velocity and acceleration is $$V = 3(t-3)^2$$ $$A = 6(t-3)$$

With domain of $t$ being $[0,5]$

All you have to do now is plug in the endpoints $0$ and $5$ into the velocity equation and whichever is the greater number would be the fastest velocity since the function is bounded, which in this case it is at time $t = 0$ and the velocity is $27$ $kph$