Question

The position of a particle moving along a line is given by s(t) = 2t3 - 24t2 + 90t + 7 for t ≥ 0. For what values of t is the speed of the particle increasing?

Answer 1

The speed of the particle increases for $t>4$

Step-by-step explanation:

Given: $s(t)=2t^{3} -24t^{2} +90t+7$

To find the speed of the particle, we need a second derivative of the equation $s(t)=2t^{3} -24t^{2} +90t+7$

Differentiating, $s(t)$ we get,

$s'(t)=6t^{2} -48t+90$

Again differentiating $s'(t)$, we get,

$s''(t)=12t-48$

Since, we need to find at what value the speed of the particle increasing, we need to equate $s''(t)=0$ to find the interval at which the speed of the particle increases. Thus,

$\begin{array}{r}{s^{\prime \prime}(t)=0} \\{12 t-48=0} \\{12(t-4)=0} \\{t-4=0} \\{t=4}\end{array}$

Thus, the interval are $(0,4]$ and $[4, \infty)$

Substituting any one value between the interval $(0,4]$ in the equation $s''(t)=12t-48$

$\begin{aligned}s^{\prime \prime}(t) &=12 t-48 \\s^{\prime \prime}(1) &=12-48 \\&=-36<0\end{aligned}$

The speed of the particle does not increase at the interval $(0,4]$

Now, substituting any one value between the interval $[4, \infty)$ in the equation $s''(t)=12t-48$

$\begin{aligned}s^{\prime \prime}(t) &=12 t-48 \\s^{\prime \prime}(5) &=60-48 \\&=12>0\end{aligned}$

Thus, the speed of the particle increases during the interval $[4, \infty)$

Hence, the speed of the particle increases for $t>4$