Q24E - University Physics with Modern Physics

Question

A pilot who accelerates at more than 4g begins to “gray out” but doesn’t completely lose consciousness. (a) Assuming constant acceleration, what is the shortest time that a jet pilot starting from rest can take to reach Mach 4 (four times the speed of sound) without graying out? (b) How far would the plane travel during this period of acceleration? (Use 331 m/s for the speed of sound in cold air.)

Step-by-Step Solution

Answer

a) the shortest time the jet pilot can take is and b) the plane will travel to a distance of $2.24\times {10}^4\hspace{0.33em}\mathrm{m}$.

Step 1: Identification of the given data

The given data can be listed below as,

 

• The pilot greys out at an acceleration of 4 g.
• The speed of the sound in the air is 331 m/s.

Step 2: Significance of Newton’s first law on the jet pilot

This law elucidates that a body will continue to be in motion unless an external force acts upon a particle.

 

The difference between the initial and the final velocity and dividing the subtraction with acceleration provides the shortest time for the jet to reach Mach 4. Furthermore, the equation of the displacement of a body gives the displacement of the plane.

Step 3: Determination of the shortest time and the displacement of the plane

a) From Newton’s first law, the velocity of the jet is expressed as:

$v=v_0+at$

Here, $v_0$ is the initial velocity which is 0 m/s and v is the final velocity which is $4\times (331m/s )=1324m/s$and a is the acceleration which is $4\hspace{0.33em}\mathrm{g}\hspace{0.33em}\mathrm{which} \mathrm{is} 39.2 \mathrm{m}/\mathrm{s}{ }^2$.

 

Substituting the values in the above expression, we get,

$$\begin{gathered} 1324\hspace{0.33em}\mathrm{m}/\mathrm{s}=0\hspace{0.33em}\mathrm{m}/\mathrm{s}+(39.2\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2)\times \mathrm{t} \\ \mathrm{t}=33.77\hspace{0.33em}\mathrm{s} \end{gathered}$$ 

Thus, the shortest time the jet pilot can take is 33.77 s.

 

b) From Newton’s first law, the displacement of the jet pilot can be expressed by:

$∆x=V_{0}t+\frac{1}{2}a{t}^2$

Here, $∆x$ is the displacement of the jet, $v_0$ is the initial velocity of the jet, t is the time taken by the jet and a is the acceleration of the jet.

 

Substituting the values in the above equation, we get-

$$\begin{gathered} \mathrm{\Delta }x=(0\mathrm{m}/\mathrm{s}\times 33.77\mathrm{s})+\frac{1}{2}\left(39.2\mathrm{m}/{\mathrm{s}}^2\right)\times (33.77\mathrm{s}{)}^2 \\ =2.24\times {10}^4\mathrm{m} \end{gathered}$$

Thus, the plane will travel a distance of $2.24\times {10}^4\mathrm{m}$.