Chapter 3.5, Problem 4QQ

a.

To determine

To find: The initial position of the particle.

Answer to Problem 4QQ

The initial position of the particle is $2$ .

Explanation of Solution

Given information: The position of particle is $s(t)=-t^2+t+2$ .

Calculation: The position of particle is $s(t)=-t^2+t+2$ .

For initial position put $t=0$ we get:

  $\begin{array}{l}s(t)=-t^2+t+2\\ $ s(0)=-{(0)}^2+0+2=2$\end{array}$

  $\begin{array}{l}s(t)=-t^2+t+2\\ $ s(0)=-{(0)}^2+0+2=2$\end{array}$

Therefore the initial position of the particle is $2$ .

b.

To determine

Find the initial velocity of the particle at time $t$ .

Answer to Problem 4QQ

Initial velocity is $-2t+1$ .

Explanation of Solution

Given information: The position of particle is $s(t)=-t^2+t+2$ .

Calculation: The position of particle is $s(t)=-t^2+t+2$ .

To find initial velocity we find the first derivative.

  $\begin{array}{l}s(t)=-t^2+t+2\\ s\text{'}(t)=-2t+1\end{array}$

Thus, initial velocity is $-2t+1$ .

c.

To determine

When the particle is moving to right.

Answer to Problem 4QQ

When the particle is moving to right then $0<t<\frac{1}{2}$ .

Explanation of Solution

Given information: Initial velocity is $-2t+1$ .

Calculation: Initial velocity is $-2t+1$ .

Particle is moving to the right so velocity is positive.

  $\begin{array}{l}-2t+1>0\\ \mathrm{subtract}1\mathrm{both}\mathrm{sides}\\ -2t+1-1>0-1\\ -2t>-1\\ \mathrm{Divide}\mathrm{by}2\mathrm{both}\mathrm{sides}\\ \frac{-2t}{2}>\frac{-1}{2}\\ -t>\frac{-1}{2}\\ t<\frac{1}{2}\end{array}$

Time is lie between $0\mathrm{and}\frac{1}{2}$ so

When the particle is moving to right then $0<t<\frac{1}{2}$ .

Draw the graph of this equation:

d.

To determine

Find the acceleration of the particle at any time $t$ .

Answer to Problem 4QQ

Acceleration is $-2$ .

Explanation of Solution

Given information: Velocity is $-2t+1$ .

Calculation: Velocity is $-2t+1$ .

Acceleration is derivative of velocity.

  $\begin{array}{l}s\text{'}(t)=-2t+1\\ s\text{'}\text{'}(t)=-2\end{array}$

Therefore acceleration is $-2$ .

e.

To determine

Find the speed of the particle at the moment when $s(t)=0$ .

Answer to Problem 4QQ

The speed of the particle at the moment when $s(t)=0$ is $2$ .

Explanation of Solution

Given information: The position of particle is $s(t)=-t^2+t+2$ .

Calculation: The position of particle is $s(t)=-t^2+t+2$ .

  $s(t)=0$ so $-t^2+t+2=0$ .

Factorize the equation we get:

  $\begin{array}{l}-t^2+t+2=0\\ -(t^2-t-2)=0\\ -(t^2-2t+t-2)=0\\ -\{t(t-2)+1(t-2)\}=0\\ -(t-2)(t+1)=0\\ -(t-2)=0\mathrm{or}t+1=0\\ -t+2=0\mathrm{or}t=-1\\ 2=t\mathrm{or}t\ne -1\end{array}$

Time cannot be negative.so $t=2$ .

Speed is absolute value of velocity.

Therefore speed is $\left|-2t+1\right|$

Put $t=2$ in $\left|-2t+1\right|$ we get:

  $\begin{array}{l}=\left|-2(2)+1\right|\\ =\left|-4+1\right|\\ =\left|-3\right|\\ =3\end{array}$

Therefore speed is $2$ .