Chapter 5, Problem 96CP

(a)

To determine

The time at which the object will be moving with speed of $15\text{m}/\text{s}$.

Answer to Problem 96CP

The object will have speed $15\text{m}/\text{s}$ after time of $3.00\text{s}$.

Explanation of Solution

The force acting ion the object provides acceleration to it and due to this acceleration; the object attains a particular velocity in a specific time.

The expression for the acceleration of the object is given as.

  $\overrightarrow{a}=\frac{\overrightarrow{F}}{m}$                                                                                                             (I)

Here, $a$ is the acceleration of the block, $F$ is the force acting on the object and $m$ is the mass of the object.

The expression for the acceleration in terms of derivative of velocity is given as.

  $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$

Rearrange the above equation for velocity as.

  ${\displaystyle \int d\overrightarrow{v}}={\displaystyle \int \overrightarrow{a}.dt}$                                                                                                  (II)

Here, $\overrightarrow{v}$ is the velocity of the object and $t$ is the time taken by the block to attain the velocity.

The expression for the magnitude of the velocity of object is given as

  $\left|\overrightarrow{v}\right|=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}$                                                                                     (III)

Here, $\left|\overrightarrow{v}\right|$ is the magnitude of velocity, $\left(v_x\right)$ is the magnitude of velocity in $\widehat{i}$-direction and $\left(v_y\right)$ is the magnitude of velocity in $\widehat{j}$-direction

Conclusion:

Substitute $\left(8.00\widehat{i}-4.00t\widehat{j}\right)$ for $\overrightarrow{F}$ and $2.00\text{kg}$ for $m$ in equation (I).

  $\begin{array}{c}\overrightarrow{a}=\frac{\left(8.00\widehat{i}-4.00t\widehat{j}\right)}{2.00\text{kg}}\\ =\left(4.00\widehat{i}-2.00t\widehat{j}\right)\end{array}$

Substitute $\left(4.00\widehat{i}-2.00t\widehat{j}\right)$ for $\overrightarrow{a}$ in equation (II).

  $\begin{array}{c}{\displaystyle \int d\overrightarrow{v}}={\displaystyle \int \left(4.00\widehat{i}-2.00t\widehat{j}\right)dt}\\ \overrightarrow{v}={\displaystyle \int \left(4.00\widehat{i}\right)dt-{\displaystyle \int \left(2.00t\widehat{j}\right)}}dt\\ =\left(4.00t\widehat{i}\right)-\left(1.00t^2\widehat{j}\right)\end{array}$

Substitute $15\text{m}/\text{s}$ for $\left|\overrightarrow{v}\right|$, $\left(4.00t\right)$ for $v_x$ and $\left(1.00t^2\right)$ for $v_y$ in equation (III).

  $\left|15\text{m}/\text{s}\right|=\sqrt{{\left(4.00t\right)}^2+{\left(1.00t^2\right)}^2}$

Take square on both the sides in above equation,

  $\begin{array}{c}{\left(4.00t\right)}^2+{\left(1.00t^2\right)}^2=225\\ 16.00t^2+1.00t^4-225=0\end{array}$

Simplify the above equation for $t$ as.

  $\begin{array}{c}\left(t^2+25\right)\left(t^2-9\right)=0\\ t=\pm 3.00\text{s}\end{array}$

The value of time cannot be negative. So, only positive value is considered.

Thus, the object will have speed $15\text{m}/\text{s}$ after time of $3.00\text{s}$.

(b)

To determine

The position of the object from initial point when speed is $15\text{m}/\text{s}$.

Answer to Problem 96CP

The object is at a distance of $20.1\text{m}$ from initial point.

Explanation of Solution

The object gets displaced by some distance when it moves under a velocity for a particular time period.

Write the expression for the position of the block as.

  $\overrightarrow{v}=\frac{d\overrightarrow{x}}{dt}$

Rearrange the above equation.

  $d\overrightarrow{x}=\overrightarrow{v}.dt$                                                                                                     (IV)

Here, $\overrightarrow{x}$ is the displacement of the particle.

Write the expression for the magnitude of the distance covered as.

  $\left|\overrightarrow{r}\right|=\sqrt{{\left(r_x\right)}^2+{\left(r_y\right)}^2}$                                                                                      (V)

Here, $\left|\overrightarrow{r}\right|$ is the magnitude of displacement, $r_x$ is the displacement in $\widehat{i}$- direction and $r_y$ is the displacement in $\widehat{j}$- direction.

Conclusion:

Substitute $\left(4.00t\widehat{i}\right)-\left(1.00t^2\widehat{j}\right)$ for $\overrightarrow{v}$ in equation (IV).

  $d\overrightarrow{x}=\left(4.00t\widehat{i}\right)-\left(1.00t^2\widehat{j}\right)dt$

Integrate the above equation.

  ${\displaystyle \int d\overrightarrow{x}}={\displaystyle \int \left(4.00t\widehat{i}\right)dt}-{\displaystyle \int \left(1.00t^2\widehat{j}\right)dt}$

Simplify the above equation.

  $\overrightarrow{x}=\left(2.00t^2\widehat{i}\right)-\left(\frac{1}{3}{t}^3\widehat{j}\right)$                                                                               (VI)

Substitute $4.00t^2$ for $r_x$, $\frac{1}{3}{t}^3$ for $r_y$ and $3.00\text{s}$ for $t$  in equation (V).

  $\begin{array}{c}\left|\overrightarrow{r}\right|=\sqrt{{\left(2.00t^2\right)}^2+{\left(\frac{1}{3}{t}^3\right)}^2}\\ =\sqrt{{\left(2.00{\left(3.00\text{s}\right)}^2\right)}^2+{\left(\frac{1}{3}{\left(3.00\text{s}\right)}^3\right)}^2}\\ =\sqrt{405}\\ =20.1\text{m}\end{array}$

Thus, the object is at a distance of $20.1\text{m}$ from initial point.

(c)

To determine

The total displacement of the object during this motion.

Answer to Problem 96CP

The displacement of the object is given as $\left(18.0\text{m}\right)\widehat{i}-\left(9.00\text{m}\right)\widehat{j}$.

Explanation of Solution

Conclusion:

Substitute $3.00\text{s}$ for $t$ in equation (VI).

  $\begin{array}{c}\overrightarrow{x}=\left(2.00{\left(3.00\text{s}\right)}^2\widehat{i}\right)-\left(\frac{1}{3}{\left(3.00\text{s}\right)}^3\widehat{j}\right)\\ =\left(18.0\text{m}\right)\widehat{i}-\left(9.00\text{m}\right)\widehat{j}\end{array}$

Thus, the displacement of the object is given as $\left(18.0\text{m}\right)\widehat{i}-\left(9.00\text{m}\right)\widehat{j}$.