Chapter 11.1, Problem 11.25P

The acceleration of a particle is defined by the relation a = −kv^2.5, where k is a constant. The particle starts at x = 0 with a velocity of 16 mm/s, and when x = 6 mm, the velocity is observed to be 4 mm/s. Determine (a) the velocity of the particle when x = 5 mm, (b) the time at which the velocity of the particle is 9 mm/s.

To determine

The velocity (v) of the particle when position (x) is $5\text{mm}$.

Answer to Problem 11.25P

The velocity (v) of the particle is $\underset{\_}{4.76\text{mm/s}}$.

Explanation of Solution

Given information:

The acceleration (a) of the particle is $-k{v}^{2.5}$.

The initial position $\left(x_0\right)$ is $0\text{m}$.

At initial velocity $\left(v_0\right)$ is $16\text{mm/s}$.

When the position (x) is $6\text{mm}$ then velocity (v) is $4\text{mm/s}$.

Calculation:

Write the relation for the acceleration (a) as given below:

$a=-k{v}^{2.5}$

Here, v is velocity and k is the constant.

Express acceleration (a) by differentiation velocity (v) with respective to position (x) below;

$a=v\frac{dv}{dx}$

Substitute $-k{v}^{2.5}$ for a.

$\begin{array}{l}-k{v}^{2.5}=v\frac{dv}{dx}\\ -k{v}^{2.5}dx=vdv\\ -kdx=v^{\left(1-2.5\right)}dv\\ -kdx=v^{\left(-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}dv\end{array}$

Apply integration.

${\displaystyle \underset{x_0=0}{\overset{x_f=x}{\int }}-kdx}={\displaystyle \underset{v_0=16}{\overset{v_f=v}{\int }}{v}^{\left(-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}dv}$

Integrate the equation.

$\begin{array}{c}{\displaystyle \underset{x_0=0}{\overset{x_f=x}{\int }}-kdx}={\displaystyle \underset{v_0=16}{\overset{v_f=v}{\int }}{v}^{\left(-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}dv}\\ {\left[-kx\right]}_0^x={\left[\frac{v^{\left(-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)+1}}{\left(-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)+1}\right]}_{16}^v\\ -kx-0={\left[-2v^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_{16}^v\end{array}$

$\begin{array}{c}-kx=-2v^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\left(-2{\left(16\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\\ kx=2v^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}\end{array}$ (1)

Calculate the value (k).

Substitute $4\text{mm/s}$ for v and $6\text{mm}$ for x in Equation (1).

$\begin{array}{l}k\left(6\right)=2{\left(4\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}\\ 6k=1-\frac{1}{2}\\ k=\frac{0.5}{6}\\ k=0.0833{\text{mm}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{s}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Calculate the velocity (v) when the position (x) is $5\text{mm}$.

Substitute $5\text{mm}$ for x and $0.0833{\text{mm}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{s}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ for k in Equation (1).

$\begin{array}{l}0.0833\left(5\right)=2v^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}\\ 0.4165+\frac{1}{2}=2v^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ v^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.9165}{2}\end{array}$

$\begin{array}{c}v={\left(0.45825\right)}^{\left(2/-1\right)}\\ v=4.76\text{mm/s}\end{array}$

Therefore, the velocity (v) of the particle is $\underset{\_}{4.76\text{mm/s}}$.

To determine

The time (t) at which the velocity (v) of the particle is $9\text{mm/s}$

Answer to Problem 11.25P

The time (t) at which the velocity (v) of the particle is $9\text{mm/s}$ is $\underset{\_}{0.171\sec }$.

Explanation of Solution

Given information:

The acceleration (a) of the particle is $-k{v}^{2.5}$.

The initial position $\left(x_0\right)$ is $0\text{m}$.

At initial velocity $\left(v_0\right)$ is $16\text{mm/s}$.

When the position (x) is $6\text{mm}$ then velocity (v) is $4\text{mm/s}$.

Calculation:

Express acceleration (a) by differentiation velocity (v) with respective to time (t) below;

$a=\frac{dv}{dt}$

Substitute $-k{v}^{2.5}$ for a.

$\begin{array}{l}-k{v}^{2.5}=\frac{dv}{dt}\\ -kdt=\frac{dv}{v^{2.5}}\\ -kdt=v^{-2.5}dv\end{array}$

Apply integration.

${\displaystyle \underset{t_0=0}{\overset{t_f=t}{\int }}-kdt}={\displaystyle \underset{v_0=16}{\overset{v_f=v}{\int }}{v}^{-2.5}dv}$

Integrate the equation.

$\begin{array}{l}{\left[-kt\right]}_0^t={\left[\frac{v^{\left(-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)+1}}{\left(-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)+1}\right]}_{16}^v\\ -kt-0={\left[-\frac{2}{3}{v}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_{16}^v\\ -kt=-\frac{2}{3}{v}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\left(-\frac{2}{3}{\left(16\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\\ kt=2v^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{96}\end{array}$ (2).

Calculate the time (t) at which the velocity (v) of the particle is $9\text{mm/s}$.

Substitute $9\text{mm/s}$ for v and $0.0833{\text{mm}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{s}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ for k in Equation (2).

$\begin{array}{l}\left(0.0833\right)t=2{\left(9\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{96}\\ t=\frac{0.01427}{0.0833}\\ t=0.171\sec \end{array}$

Therefore, the time (t) at which the velocity (v) of the particle is $9\text{mm/s}$ is $\underset{\_}{0.171\sec }$.