Chapter 2, Problem 50P

(a)

To determine

The acceleration of the sneeze as it moves the first $0.25\text{ cm}$ .

Answer to Problem 50P

The acceleration of the sneeze as it moves the first $0.25\text{ cm}$ is $3.9\times {10}^5{\text{ m/s}}^2$ .

Explanation of Solution

Write the equation of motion.

$v_{fx}^2-v_{ix}^2=2a_x\Delta x$

Here, $v_{fx}$ is the final speed, $v_{ix}$ is the initial speed, $a_x$ is the acceleration and $\Delta x$ is the displacement

Rewrite the above equation for $a_x$ .

$a_x=\frac{v_{fx}^2-v_{ix}^2}{2\Delta x}$ (I)

Conclusion:

Given that the maximum speed of the sneeze is $44\text{ m/s}$ and its initial speed is $0$ .

Substitute $0.25\text{ cm}$ for $\Delta x$ , $44\text{ m/s}$ for $v_{fx}$ and $0$ for $v_{ix}$ in equation (I) to find the value of $a_x$ .

$\begin{array}{c}{a}_x=\frac{{\left(44\text{ m/s}\right)}^2-0}{2\left(0.25\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}\\ =\frac{{\left(44\text{ m/s}\right)}^2}{2\left(0.0025\text{ m}\right)}\\ =3.9\times {10}^5{\text{ m/s}}^2\end{array}$

Therefore, the acceleration of the sneeze as it moves the first $0.25\text{ cm}$ is $3.9\times {10}^5{\text{ m/s}}^2$ .

(b)

To determine

The time taken by the sneeze to move the $2.0\text{ cm}$ distance in the nose.

Answer to Problem 50P

The time taken by the sneeze to move the $2.0\text{ cm}$ distance in the nose is $0.51\text{ ms}$ .

Explanation of Solution

Write the equation connecting displacement and time.

$\Delta x=\frac{1}{2}\left(v_{fx}+v_{ix}\right)\Delta t$

Here, $\Delta t$ is the time taken for the displacement

Rewrite the above equation for $\Delta t$ and use this equation to find the time taken for first $0.25\text{ cm}$ displacement.

$\Delta t_1=\frac{2\Delta x}{v_{fx}+v_{ix}}$ (II)

Here, $\Delta t_1$ is the time taken for first $0.25\text{ cm}$ displacement

Write the equation of motion.

$\Delta x=v_{ix}\Delta t+\frac{1}{2}{a}_x{\left(\Delta t\right)}^2$

Use the above equation to find the time taken for the displacement when the sneeze moves with constant velocity. Since the velocity is constant, the acceleration will be zero.

$\begin{array}{c}\Delta x=v_{ix}\Delta t+0\\ =v_{ix}\Delta t_2\end{array}$

Here, $\Delta t_2$ is the time taken for the displacement after $0.25\text{ cm}$

Rewrite the above equation for $\Delta t_2$ .

$\Delta t_2=\frac{\Delta x}{v_{ix}}$ (III)

Write the equation for the time taken to cover the total distance.

$\Delta t=\Delta t_1+\Delta t_2$ (IV)

Conclusion:

Substitute $0.25\text{ cm}$ for $\Delta x$ , $44\text{ m/s}$ for $v_{fx}$ and $0$ for $v_{ix}$ in equation (II) to find the value of $\Delta t_1$ .

$\begin{array}{c}\Delta t_1=\frac{2\left(0.25\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}{44\text{ m/s}+0}\\ =\frac{2\left(0.0025\text{ m}\right)}{44\text{ m/s}}\\ =0.11\times {10}^{-3}\text{ s}\\ =0.11\text{ ms}\end{array}$

Substitute $1.75\text{ cm}$ for $\Delta x$ and $44\text{ m/s}$ for $v_{ix}$ in equation (III) to find $\Delta t_2$ .

$\begin{array}{c}\Delta t_2=\frac{1.75\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}{44\text{ m/s}}\\ =\frac{0.0175\text{ m}}{44\text{ m/s}}\\ =0.4\times {10}^{-3}\text{ s}\\ =0.4\text{ ms}\end{array}$

Substitute $0.11\text{ ms}$ for $\Delta t_1$ and $0.40\text{ ms}$ for $\Delta t_2$ in equation (IV) to find $\Delta t$ .

$\begin{array}{c}\Delta t=0.11\text{ ms}+0.40\text{ ms}\\ =0.51\text{ ms}\end{array}$

Therefore, the time taken by the sneeze to move the $2.0\text{ cm}$ distance in the nose is $0.51\text{ ms}$ .

(c)

To determine

The graph of $v_x\left(t\right)$ .

Answer to Problem 50P

The graph of $v_x\left(t\right)$ is

Explanation of Solution

Velocity is the rate of change of displacement of the object. In velocity-time graph, the instantaneous value of the velocity of an object is plotted as a function of time. The slope of the velocity-time graph gives the acceleration of the motion while the area under the graph gives the displacement during the motion.

The $v_x\left(t\right)$ graph of the sneeze is shown in figure 1.

The initial velocity of the sneeze is $0$ and the velocity increases up to $0.11\text{ ms}$ to the maximum value of $44\text{ m/s}$ . The sneeze then moves with constant velocity till $0.51\text{ ms}$ .