Question

A particle moves with position function

\(s = {t^4} - 4{t^3} - 20{t^2} + 20t\), \(t > 0\)

(a) At what time does the particle have a velocity of 20 m/s?

(b) At what time is the acceleration 0? What is the significance of this value of \(t\)?

Short answer

a). The velocity is 20 m/s at 0 s and 5 s.

b). The acceleration is 0 at \(3.08\) s. At this value of time, acceleration is shifting from negative value to positive value, which corresponds to minimum velocity point.

Step-by-step solution

(a) Step 1: Analyze the duration of positive velocity

The velocity is obtained as the rate of change of the distance, which is \(v\left( t \right) = s'\left( t \right)\) .

Thus, \(v\left( t \right) = 4{t^3} - 12{t^2} - 40t + 20\). Plug \(v = 20\) and solve for time, as shown below:

\(\begin{aligned}4{t^3} - 12{t^2} - 40t + 20 &= 20\\4{t^3} - 12{t^2} - 40t &= 0\\4t\left( {{t^2} - 3t - 10} \right) &= 0\\4t\left( {t - 5} \right)\left( {t + 2} \right) &= 0\\t &= 0,\,5\end{aligned}\)

It is to be noted that we consider only positive values of time as \(t > 0\).

(b) Step 2: Find the acceleration

The acceleration is the rate of change of velocity \(a\left( t \right) = v'\left( t \right)\) and is equal to \(a\left( t \right) = 12{t^2} - 24t - 40\) . Set \(a\left( t \right) = 0\) and solve for the time \(t\), at which the acceleration is 0, as follows:

\(\begin{aligned}12{t^2} - 24t - 40 &= 0\\4\left( {3{t^2} - 6t - 10} \right) &= 0\\t &= \frac{{6 \pm \sqrt {{6^2} - 4\left( 3 \right)\left( { - 10} \right)} }}{{2\left( 3 \right)}}\\ &= 1 \pm \frac{1}{3}\sqrt {39} \\ \approx 3.08\,{\rm{s}}\end{aligned}\)

It is to be noted that we consider only positive values of time as \(t > 0\). Moreover, at this value of time, acceleration is shifting from negative value to positive value, which corresponds to minimum velocity point.