Q25E - University Physics with Modern Physics

Question

A jet plane comes in for a downward dive as shown in Fig. E3.25. The bottom part of the path is a quarter circle with a radius of curvature of 280 m. According to medical tests, pilots will lose consciousness when they pull out of a dive at an upward acceleration greater than 5.5g. At what speed (in m/s and in mph) will the pilot black out during this dive?

Step-by-Step Solution

Answer

The velocity of pilot during the dive is 128.8 m/s or 274.7 mph.

Step 1: Identification of given data:

The given data can be listed below,

• The radius of curvature of the path is, R = 280 m
• The radial acceleration of the pilots is, $a_r=5.5g=5.5\times 9.8m/s^2=53.9m/s^2$.

Step 2: Concept/Significance of radial velocity:

A particle's velocity along the observer's line of sight is known as its radial component. It is a part of a radial direction change in location per unit of time.

Step 3: Determination of the speed (in m/s and in mph) will the pilot blackout during the dive:

The velocity of the pilot is given by,

$$\begin{gathered} a_r=\frac{v^2}{R} \\ v=\sqrt{a_{r}R} \end{gathered}$$

Here, a_r is the radial acceleration, and R is the radius of the path.

Substitute all the values in the above expression.

$\begin{array}{rcl}v& =& \sqrt{\left(53.9m/s^2\right)\left(280m\right)}\\ & =& 122.8m/s\left(\frac{1mph}{0.4770m/s}\right)\\ & =& 274.7mph\end{array}$

Hence, the velocity of pilot during the dive is 128.8 m/s or 274.7 mph.