Chapter 4.1, Problem 63E

To determine

To find: the velocity and acceleration of the particle moving along a coordinate line is $s=\sqrt{1+4t}$ at $t=6\text{seconds}$ .

Answer to Problem 63E

The velocity of the particle is $v\left(6\right)=\frac{2}{5}\text{m}/\text{s}$ .

The acceleration of the particle is $a\left(6\right)=-\frac{4}{125}\text{m}/\text{s}$

Explanation of Solution

Given information:

A particle moving along the line $s=\sqrt{1+4t}$

  $s$ in meters and $t$ in seconds.

Calculation:

The first derivative of the line $s=\sqrt{1+4t}$ gives the velocity of the particle,

  $\begin{array}{c}{s}^{\prime }=\frac{d}{dt}\left(\sqrt{1+4t}\right)\\ s^{\prime }=\frac{d}{dt}{\left(1+4t\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ s^{\prime }=\frac{1}{2}{\left(1+4t\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times 4\\ v=\frac{2}{\sqrt{1+4t}}\end{array}$

Velocity at $t=6\text{seconds}$ ,

  $\begin{array}{l}v\left(6\right)=\frac{2}{\sqrt{1+4\left(6\right)}}\\ v\left(6\right)=\frac{2}{\sqrt{25}}\\ v\left(6\right)=\frac{2}{5}\text{m}/\text{s}\end{array}$

The second derivative of the line $s=\sqrt{1+4t}$ gives the acceleration of the particle,

  $\begin{array}{c}{s}^{″}=\frac{d}{dt}\left(\frac{2}{\sqrt{1+4t}}\right)\\ s^{″}=\frac{d}{dt}\left(2{\left(1+4t\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\\ s^{″}=2\left(-\frac{1}{2}\right){\left(1+4t\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ a=-\frac{4}{{\left(1+4t\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}$

Acceleration at $t=6\text{seconds}$ ,

  $\begin{array}{l}a\left(6\right)=-\frac{4}{{\left(1+4\left(6\right)\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ a\left(6\right)=-\frac{4}{{\left(25\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ a\left(6\right)=-\frac{4}{125}\text{m}/\text{s}\end{array}$

Therefore we can conclude that velocity at $t=6\text{seconds}$ is $v\left(6\right)=\frac{2}{5}\text{m}/\text{s}$ and acceleration at $t=6\text{seconds}$ is $a\left(6\right)=-\frac{4}{125}\text{m}/\text{s}$ .