Chapter 8, Problem 4P

To determine

Find the velocity and position of the particle at $t=4\text{s}$.

Answer to Problem 4P

The velocity and position of the particle at $t=4\text{s}$ is $\underset{\_}{2000.6a_x+5\times {10}^3{a}_y\text{m/s}}$ and $\underset{\_}{\left(8001.2,20000,0\right)}$ respectively.

Explanation of Solution

Calculation:

Write the expression for Newton’s second law of motion.

$\begin{array}{c}F=ma\\ =QE\end{array}$

Rearrange the above expression as follows.

$a=\frac{QE}{m}$

Substitute 2kg for m, $30a_x\text{kV/m}$ for E and $10\text{ mC}$ for Q to find acceleration (a).

$\begin{array}{c}a=\frac{\left(10\text{ mC}\right)\left(30a_x\text{kV/m}\right)}{2\text{kg}}\\ =0.15a_x{\text{m/s}}^{\text{2}}\end{array}$

Write the expression for acceleration.

$\begin{array}{c}a=\frac{du}{dt}\\ =\frac{d}{dt}\left(u_x,u_y,u_z\right)\end{array}$

Substitute $0.15a_x{\text{m/s}}^{\text{2}}$ for a.

$\frac{d}{dt}\left(u_x,u_y,u_z\right)=0.15a_x$

Equate and integrate the above equation to find $u_x$, $u_y$ and $u_z$.

$\begin{array}{l}\frac{d{u}_x}{dt}=0.15\\ {\displaystyle \int d{u}_x={\displaystyle \int 0.15dt}}\end{array}$

$u_x=0.15t+A$        (1)

Here,

A is the integration constant.

$\begin{array}{l}\frac{d{u}_y}{dt}=0\\ {\displaystyle \int d{u}_y={\displaystyle \int 0dt}}\end{array}$

$u_y=B$        (2)

Here,

B is the integration constant.

$\begin{array}{l}\frac{d{u}_z}{dt}=0\\ {\displaystyle \int d{u}_z={\displaystyle \int 0dt}}\end{array}$

$u_z=C$        (3)

Here,

C is the integration constant.

At $t=0$, the given value of velocity is $u=\left(2a_x+5a_y\right)\times {10}^3\text{m/s}$.

Modify equation (1) as follows.

$u_x\left(t=0\right)=0.15t+A$

Substitute 0 for t and $2\times {10}^3$ for $u_x\left(t=0\right)$ to find the value of A.

$\begin{array}{l}2\times {10}^3=0.15\left(0\right)+A\\ A=2\times {10}^3\end{array}$

Modify equation (2) as follows.

$u_y\left(t=0\right)=B$

Substitute $5\times {10}^3$ for $u_y\left(t=0\right)$ to find the value of B.

$5\times {10}^3=B$

Modify equation (3) as follows.

$u_z\left(t=0\right)=C$

Substitute $0$ for $u_z\left(t=0\right)$ to find the value of C.

$0=C$

Substitute $2\times {10}^3$ for A in equation (1).

$u_x=0.15t+2\times {10}^3$

Substitute $5\times {10}^3$ for B in equation (2).

$u_y=5\times {10}^3$

Substitute $0$ for C in equation (3).

$u_z=0$

Represent u(t) in terms of $u_x$, $u_y$ and $u_z$.

$\begin{array}{c}u\left(t\right)=\left(u_x,u_y,u_z\right)\\ =\left(0.15t+2\times {10}^3,5\times {10}^3,0\right)\end{array}$

Find the value of $u\left(t=4\text{s}\right)$.

$\begin{array}{c}u\left(t=4\text{s}\right)=\left(\left(0.15\left(4\right)+2\times {10}^3\right)a_x+5\times {10}^3{a}_y+0a_z\right)\\ =2000.6a_x+5\times {10}^3{a}_y\text{m/s}\end{array}$

Write the expression for velocity.

$\begin{array}{c}u=\frac{dl}{dt}\\ =\frac{d}{dt}\left(x,y,z\right)\end{array}$

Substitute $\left(0.15t+2\times {10}^3,5\times {10}^3,0\right)$ for u.

$\left(0.15t+2\times {10}^3,5\times {10}^3,0\right)=\frac{d}{dt}\left(x,y,z\right)$

Equate and integrate the above equation to find $x$, $y$ and $z$.

$\begin{array}{l}\frac{dx}{dt}=0.15t+2\times {10}^3\\ {\displaystyle \int dx={\displaystyle \int \left(0.15t+2\times {10}^3\right)dt}}\\ x=\frac{0.15t^2}{2}+2\times {10}^{3}t+A_1\end{array}$

$x=0.075t^2+2\times {10}^{3}t+A_1$        (4)

Here,

$A_1$ is the integration constant.

$\begin{array}{l}\frac{dy}{dt}=5\times {10}^3\\ {\displaystyle \int dy={\displaystyle \int \left(5\times {10}^3\right)dt}}\end{array}$

$y=5\times {10}^{3}t+B_1$        (5)

Here,

$B_1$ is the integration constant.

$\begin{array}{l}\frac{dz}{dt}=0\\ {\displaystyle \int dz={\displaystyle \int 0dt}}\end{array}$

$z=C_1$        (6)

Here,

$C_1$ is the integration constant.

Since, the particle starts at origin, at $t=0$ the value of $\left(x,y,z\right)=\left(0,0,0\right)$.

Modify equation (4) as follows.

$x\left(t=0\right)=0.075t^2+2\times {10}^{3}t+A_1$

Substitute 0 for t and $0$ for $x\left(t=0\right)$ to find the value of $A_1$.

$\begin{array}{l}0=0.075{\left(0\right)}^2+2\times {10}^3\left(0\right)+A_1\\ A_1=0\end{array}$

Modify equation (5) as follows.

$y\left(t=0\right)=5\times {10}^{3}t+B_1$

Substitute 0 for t and $0$ for $y\left(t=0\right)$ to find the value of $B_1$.

$\begin{array}{l}0=5\times {10}^3\left(0\right)+B_1\\ B_1=0\end{array}$

Modify equation (6) as follows.

$z\left(t=0\right)=C_1$

Substitute $0$ for $z\left(t=0\right)$ to find the value of $C_1$.

$0=C_1$

Substitute 0 for $A_1$ in equation (4).

$\begin{array}{c}x=0.075t^2+2\times {10}^{3}t+0\\ =0.075t^2+2\times {10}^{3}t\end{array}$

Substitute 0 for $B_1$ in equation (5).

$\begin{array}{c}y=5\times {10}^{3}t+0\\ =5\times {10}^{3}t\end{array}$

Substitute 0 for $C_1$ in equation (6).

$z=0$

Represent the points $\left(x,y,z\right)$ in terms of t as follows.

$\left(x,y,z\right)=\left(0.075t^2+2\times {10}^{3}t,5\times {10}^{3}t,0\right)$

Find the value of $\left(x,y,z\right)$ at $t=4\text{s}$.

$\begin{array}{c}\left(x,y,z\right)=\left(0.075{\left(4\right)}^2+2\times {10}^3\left(4\right),5\times {10}^3\left(4\right),0\right)\\ =\left(8001.2,20000,0\right)\end{array}$

Conclusion:

Thus, the velocity and position of the particle at $t=4\text{s}$ is $\underset{\_}{2000.6a_x+5\times {10}^3{a}_y\text{m/s}}$ and $\underset{\_}{\left(8001.2,20000,0\right)}$ respectively.