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A jet fighter pilot wishes to accelerate from rest at a constant acceleration of 5 g to reach Mach 3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.0 s. Use 331 m/s for the speed of sound. (a) Will the period of acceleration last long enough to cause him to black out? (b) What is the greatest speed he can reach with an acceleration of 5g before he blacks out?

User NIcE COw
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2 Answers

4 votes

Final answer:

The jet fighter pilot will black out during the acceleration period as it lasts longer than 5.0 s. The greatest speed the pilot can reach with an acceleration of 5g before blacking out is 245 m/s.

Step-by-step explanation:

To determine whether the jet fighter pilot will black out, we need to calculate the time of acceleration. The speed of sound is v = 331 m/s. The speed the pilot wants to reach is 3 times the speed of sound, so the desired speed is 3 * 331 m/s = 993 m/s. We can use the equation:

v = u + at

Where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration (5 g = 5 * 9.8 m/s^2 = 49 m/s^2), and t is the time.

Substituting the values, we get:

993 m/s = 0 m/s + 49 m/s² x t

Rearranging this equation to solve for t, we get:

t = 993 m/s / 49 m/s^2 = 20.265 s

Since the acceleration lasts for longer than 5.0 seconds, the pilot will black out during the acceleration period.

To calculate the greatest speed the pilot can reach before blacking out, we can use the same equation, but rearrange it to solve for v:

v = u + at

Substituting the values, we get:

v = 0 m/s + 49 m/s^2 x 5.0 s = 245 m/s

Therefore, the greatest speed the pilot can reach with an acceleration of 5 g before blacking out is 245 m/s.

User Esteve
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6.2k points
6 votes

Answer:

20.244648318 s

10051.4678899 m

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity =
3* 331

s = Displacement

a = Acceleration =
5g=5* 9.81\ m/s^2

g = Acceleration due to gravity = 9.81 m/s²


v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(3* 331-0)/(5* 9.81)\\\Rightarrow t=20.244648318\ s

The time taken is 20.244648318 s


s=ut+(1)/(2)at^2\\\Rightarrow s=0* t+(1)/(2)* 5* 9.81* 20.244648318^2\\\Rightarrow s=10051.4678899\ m

The plane would travel 10051.4678899 m

User Parag
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6.0k points