Question

The speed required of a military jet when taking off from the deck of an aircraft carrier is dependent upon the speed of the carrier and the speed of the wind into which the carrier is moving. The takeoff speed required of a military jet relative to the deck of the carrier is 45 m/s when the carrier travels at 45 mi/hr into a 20 mi/hr wind. And when the aircraft carrier is traveling at 10 mi/hr into a 5 mi/hr wind, the takeoff speed relative to the deck of the carrier is 71 m/s. Determine the acceleration which a military jet must have to take off under these two conditions from the 126-m long runway of the USS Ronald Reagan aircraft carrier.

Answer 1

Answer:

Explanation:

Given

Velocity of jet relative to the carrier$(v_{jc})=45 m/s$

Velocity of jet carrier to the wind$(v_{cw})=45 mi/hr\approx 20.1168$

Velocity of wind$=20 mi/hr\approx 8.9408 m/s$

Length of runway=126 m

$v_{jc}=v_j-v_c=45$

$v_{cw}=v_c-v_w=20.11$

$v_w=8.94 m/s$

thus $v_c$=20.11+8.94=29.05

$v_j$=29.05+45=74.05 m/s

to find acceleration

$v^2-u^2=2as$

here u=0,v=74.05 m/s

$a=\frac{v^2}{2s}=\frac{74.05^2}{2\times 126}=21.75 m/s^2$

case -2

Velocity of jet relative to the carrier$(v_{jc})=71 m/s$

Velocity of jet carrier to the wind$(v_{cw})=10 mi/hr\approx 4.4704$

Velocity of wind$=5 mi/hr\approx 2.2352 m/s$

Length of runway=126 m

$v^2-u^2=2as$

here u=0,v=77.7 m/s

$a=\frac{v^2}{2s}=\frac{77.7^2}{2\times 126}=23.95 m/s^2$