If a rocket with a mass of 3500 tons vertically accelerates at a rate of # 3/5 m/s^2#, how much power will the rocket have to exert to maintain its acceleration at 12 seconds?

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Answer 1 · 2016-04-17T12:45:46

#15.12MW#

Velocity of the rocket at #12s# is

#v=u+at#
#v=0m/s+3/5m/s^2*12s=7.2m/s#

Power is #"work"/"time"=W/t#, and work is #"force"*"distance"=Fd#, so

#P=W/t=(Fd)/t#

and, because velocity is distance over time, #d/t#,

#P=Fv#

#F# is found by Newtons second law,

#F=ma#
#F=3,500,000kg*3/5m/s^2=2,100,000N#

so

#P=2,100,000N*7.2m/s=15,120,000Nm/s=15,120,000W#

#P=15.12MW#