Chapter 2.6, Problem 15E

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s = 1/t², where t is measured in seconds. Find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.

To determine

To find: The velocity of the particle when $t=a,t=1,t=2$ and $t=3$ seconds.

Answer to Problem 15E

The velocity of the particle at times $t=a,t=1,t=2$ and $t=3$ are $\underset{\_}{-\frac{2}{a^3}\text{ m/s}}$, $\underset{\_}{-2\text{ m/s}}$ , $\underset{\_}{-\frac{1}{4}\text{ m/s}}$ and $\underset{\_}{-\frac{2}{27}\text{ m/s}}$, respectively.

Explanation of Solution

Given:

The displacement (in meters) of a particle moving in a straight line is $s=\frac{1}{t^2}$, where t is measured in seconds.

Formula used:

The derivative $v\left(a\right)$ is the velocity of the particle at time $t=a$,

$v\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ (1)

Calculation:

Obtain the velocity of the particle at $t=a$.

Take the position function $s\left(t\right)=f\left(t\right)$ and use equation (1) to compute the velocity of the particle at time $t=a$ as follows.

$\begin{array}{c}v\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(\frac{1}{{\left(a+h\right)}^2}\right)-\left(\frac{1}{{\left(a\right)}^2}\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(\frac{1}{\left(a^2+h^2+2ah\right)}\right)-\left(\frac{1}{a^2}\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(\frac{a^2-\left(a^2+h^2+2ah\right)}{a^2\left(a^2+h^2+2ah\right)}\right)}{h}\end{array}$

Simplify further and obtain the velocity as follows.

$\begin{array}{c}v\left(a\right)=\underset{h\to 0}{\lim }\frac{-\left(h^2+2ah\right)}{h{a}^2\left(a^2+h^2+2ah\right)}\\ =-\underset{h\to 0}{\lim }\frac{h\left(h+2a\right)}{h{a}^2\left(a^2+h^2+2ah\right)}\end{array}$

Cancel the common term h from both the numerator and the denominator,

$\begin{array}{c}v\left(a\right)=-\underset{h\to 0}{\lim }\frac{\left(h+2a\right)}{a^2\left(a^2+h^2+2ah\right)}\\ =-\frac{\left(0+2a\right)}{a^2\left(a^2+{\left(0\right)}^2+2a\left(0\right)\right)}\\ =-\frac{2a}{a^2\left(a^2\right)}\\ =-\frac{2}{a^3}\end{array}$

Thus, the velocity of the particle at time $t=a$ is $\underset{\_}{-\frac{2}{a^3}\text{ m/s}}$.

Obtain the velocity of the particle at time $t=1$ second.

Substitute $a=1\text{ s}$ in $v\left(a\right)$,

$\begin{array}{c}v\left(1\right)=-\frac{2}{{\left(1\right)}^3}\\ =-2\end{array}$

Thus, the velocity of the particle at time $t=1\text{ s}$ is $\underset{\_}{-2\text{ m/s}}$.

Obtain the velocity of the particle at time $t=2$ seconds.

Substitute $a=2\text{ s}$ in $v\left(a\right)$,

$\begin{array}{c}v\left(2\right)=-\frac{2}{{\left(2\right)}^3}\\ =-\frac{2}{8}\\ =-\frac{1}{4}\end{array}$

Thus, the velocity of the particle at time $t=2\text{ s}$ is $\underset{\_}{-\frac{1}{4}\text{ m/s}}$ .

Obtain the velocity of the particle at time $t=3$ seconds.

Substitute $a=3\text{ s}$ in $v\left(a\right)$,

$\begin{array}{c}v\left(3\right)=-\frac{2}{{\left(3\right)}^3}\\ =-\frac{2}{27}\end{array}$

Thus, the velocity of the particle at time $t=3\text{ s}$ is $\underset{\_}{-\frac{2}{27}\text{ m/s}}$.