Chapter 5, Problem 5.96CP

A time-dependent force, $\stackrel{\to }{\text{F}}$ = (8.00 $\hat{\text{i}}$ - 4.00/ $\hat{\text{j}}$), where $\stackrel{\to }{\text{F}}$ is in newtons and t is in seconds, is exerted on a 2.00-kg object initially at rest, (a) At what time will the object be moving with a speed of 15.0 m/s? (b) How far is the object from its initial position when its speed is 15.0 m/s? (c) Through what total displacement has the object traveled at this moment?

To determine

The time for which the object moves with a speed of $15.0\text{m}/\text{s}$

Answer to Problem 5.96CP

The time taken by the object is $3.00\text{s}$.

Explanation of Solution

Given info: The time-dependent force vector is $\left(8.00\widehat{\text{i}}-4.00t\widehat{\text{j}}\right)\text{N}$ and the mass of an object initially at rest is $2.00\text{kg}$.

From the Newton’s law of motion, the force on the object is,

Here,

$m$ is the mass of an object.

is the acceleration of an object.

Rearrange the above formula.

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

The expression for the acceleration is,

Substitute $\left(4\text{m}/{\text{s}}^2\right)\widehat{\text{i}}-\left(2t\text{m}/{\text{s}}^2\right)\widehat{\text{j}}$ for in the above equation.

(1)

The expression for the velocity vector is,

Here,

$v_{\text{x}}$ is the $x$-component of the velocity vector.

$v_{\text{y}}$ is the $y$-component of the velocity vector.

$v_{\text{z}}$ is the $z$-component of the velocity vector.

Compare the above equation with equation (1), the $x$-component of the velocity is $4t$, $y$-component of the velocity is $t^2$ and $z$-component of the velocity is zero.

The magnitude of the velocity vector is,

$v=\sqrt{v_{\text{x}}^2+v_{\text{y}}^2+v_{\text{z}}^2}$

Substitute $4t$ for $v_{\text{x}}$, $t^2$ for $v_{\text{y}}$ and $0$ for $v_{\text{z}}$ in the above equation.

$\begin{array}{c}v=\sqrt{{\left(4t\right)}^2+{\left(t^2\right)}^2+{\left(0\right)}^2}\\ =\sqrt{16t^2+t^4}\end{array}$

Substitute $15.0\text{m}/\text{s}$ for $v$ in the above equation.

$\begin{array}{c}15.0\text{m}/\text{s}=\sqrt{16t^2+t^4}\\ 16t^2+t^4=225\\ t^4+16t^2-225=0\\ t^4+25t^2-9t^2-225=0\end{array}$

Further solve the above equation.

$\begin{array}{c}\left(t^4+25t^2\right)-\left(9t^2+225\right)=0\\ t^2\left(t^2+25\right)-9\left(t^2+25\right)=0\\ \left(t^2-9\right)\left(t^2+25\right)=0\end{array}$

In the above fraction of equation, second factor give the negative value of time and time cannot be negative so the value of time from first factor is,

$\begin{array}{c}{t}^2-9=0\\ t^2=9\\ t=3.00\text{s}\end{array}$

Conclusion:

Therefore, the time taken by the object is $3.00\text{s}$.

To determine

The position of an object from its initial position.

Answer to Problem 5.96CP

The position of an object from its initial position is $20.1\text{m}$.

Explanation of Solution

From part (a), the expression for the velocity of the object is,

The expression for the velocity of the object is,

Here,

is the position vector of the object.

Substitute $\left(4t\widehat{\text{i}}-t^2\widehat{\text{j}}\right)\text{m}/\text{s}$ for in the above equation.

(2)

The expression for the position vector from its initial position is,

Here,

$r_{\text{x}}$ is the $x$-component of the position vector.

$r_{\text{y}}$ is the $y$-component of the position vector.

$r_{\text{z}}$ is the $z$-component of the position vector.

Compare the above equation with equation (1), the $x$-component of the position is $2t^2$, $y$-component of the position is $\frac{t^3}{3}$ and $z$-component of the position is zero.

The magnitude of the velocity vector is,

$r=\sqrt{r_{\text{x}}^2+r_{\text{y}}^2+r_{\text{z}}^2}$

Substitute $2t^2$ for $r_{\text{x}}$, $\frac{t^3}{3}$ for $r_{\text{y}}$ and $0$ for $r_{\text{z}}$ in the above equation.

$\begin{array}{c}r=\sqrt{{\left(2t^2\right)}^2+{\left(\frac{t^3}{3}\right)}^2+{\left(0\right)}^2}\\ =\sqrt{4t^4+\frac{t^6}{9}}\end{array}$

From part (a), the time taken by the object is,

$t=3.00\text{s}$

Substitute $3.00\text{s}$ for $t$ in the above equation.

$\begin{array}{c}r=\sqrt{4{\left(3.00\text{s}\right)}^4+\frac{{\left(3.00\text{s}\right)}^6}{9}}\text{m}\\ \approx 20.1\text{m}\end{array}$

Conclusion:

Therefore, the position of an object from its initial position is $20.1\text{m}$.

To determine

The total displacement of the object.

Answer to Problem 5.96CP

The total displacement of the object is $\left(18.0\widehat{\text{i}}-9.0\widehat{\text{j}}\right)\text{m}$.

Explanation of Solution

From part (b), the position vector of the object is,

From part (a), the time taken by the object is,

$t=3.00\text{s}$

Substitute $3.00\text{s}$ for $t$ in the above equation to find the displacement of the object.

Conclusion:

Therefore, the total displacement of the object is $\left(18.0\widehat{\text{i}}-9.0\widehat{\text{j}}\right)\text{m}$.