Chapter 2.4, Problem 19E

a.

To determine

To calculate: The displacement of the particle.

Answer to Problem 19E

The displacement of the particle is $2$ m.

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^2-3t+2$ and the initial time is $t=0$ and final time is $t=5$

Concept used:

The displacement of the particle is final distance minus initial distance.

Calculation:

Substitute $5$ for $t$ in $s\left(t\right)=t^2-3t+2$ .

  $\begin{array}{c}s\left(5\right)={\left(5\right)}^2-3\left(5\right)+2\\ =25-15+2\\ =27-15\\ =12\end{array}$

Substitute $0$ for $t$ in $s\left(t\right)=t^2-3t+2$ .

  $\begin{array}{c}s\left(0\right)={\left(0\right)}^2-3\left(0\right)+2\\ =0-0+2\\ =2\end{array}$

Subtract initial distance from the final distance for the displacement of the particle.

Displacement is $12-10=2$

Conclusion: The displacement of the particle is $2$ m.

b.

To determine

To calculate: The average velocity of the particle.

Answer to Problem 19E

The average velocity of the particle is $2m/s$ .

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^2-3t+2$ and the initial time is $t=0$ and final time is $t=5$

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ , $\text{Average velocity = }\frac{\text{displacement}}{\text{Change in time}}$

Calculation:

The function is $s\left(t\right)=t^2-3t+2$ .

Displacement is $12-10=2$

  $\begin{array}{c}\text{Average velocity = }\frac{10}{5-0}\\ =\frac{10}{5}\\ =2\end{array}$

Conclusion: The average velocity is $2m/s$

c.

To determine

To calculate: The instantaneous velocity.

Answer to Problem 19E

The instantaneous velocity is $5$ m/s.

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^2-3t+2$ and time is $t=4$

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

Calculation:

The function is $s\left(t\right)=t^2-3t+2$ .

Differentiate the function with respect to time $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left(t^2-3t+1\right)\\ =\frac{d}{dt}\left(t^2\right)-3\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(1\right)\\ =2t-3\left(1\right)+0\\ =2t-3\end{array}$

Substitute $4$ for $t$ in $\frac{ds}{dt}=2t-3$ .

  $\begin{array}{c}\frac{ds}{dt}=2\times 4-3\\ =8-3\\ =5\end{array}$

Conclusion: The instantaneous velocity of the particle is $5$ m/s.

d.

To determine

To calculate: The acceleration of the particle.

Answer to Problem 19E

The acceleration of the particle is $2$

  $m/s^2$ .

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^2-3t+2$ and time is $t=4$

Calculation:

The function is $s\left(t\right)=t^2-3t+2$ .

Differentiate the function with respect to time $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left(t^2-3t+1\right)\\ =\frac{d}{dt}\left(t^2\right)-3\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(1\right)\\ =2t-3\left(1\right)+0\\ =2t-3\end{array}$

Differentiate the function again with respect to time $t$ .

  $\begin{array}{c}\frac{d}{dt}\left(\frac{ds}{dt}\right)=\frac{d}{dt}\left(2t-3\right)\\ =\frac{d}{dt}\left(2t\right)-3\frac{d}{dt}\left(1\right)\\ =2-0\\ =2\end{array}$

Conclusion: The acceleration of the particle is $2$ $m/s^2$ .

e.

To determine

To calculate: The acceleration of the particle is $2$

  $m/s^2$ .

Answer to Problem 19E

The acceleration of the particle is

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^2-3t+2$ and time is $t=4$

Calculation:

The function is $s\left(t\right)=t^2-3t+2$ .

Differentiate the function with respect to time $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left(t^2-3t+1\right)\\ =\frac{d}{dt}\left(t^2\right)-3\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(1\right)\\ =2t-3\left(1\right)+0\\ =2t-3\end{array}$

Differentiate the function again with respect to time $t$ .

  $\begin{array}{c}\frac{d}{dt}\left(\frac{ds}{dt}\right)=\frac{d}{dt}\left(2t-3\right)\\ =\frac{d}{dt}\left(2t\right)-3\frac{d}{dt}\left(1\right)\\ =2-0\\ =2\end{array}$

Conclusion: The acceleration of the particle is $2$ $m/s^2$ .

f.

To determine

To calculate: The time from the function starts increasing.

Answer to Problem 19E

The function starts increasing from $t=\frac{3}{2}$ sec.

Explanation of Solution

Given information:

The function is $s\left(t\right)=t^2-3t+2$ and time is $t=4$

Calculation:

The function is $s\left(t\right)=t^2-3t+2$ .

The graph of the function is shown below.

  

From the graph of the function, it is clear that the function increases from $t=\frac{3}{2}$ sec.