If a rocket with a mass of #2500# tons vertically accelerates at a rate of # 7/3 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 · 2016-04-17T12:53:03

#P=163 1/3MW#

The velocity at #12s# is

#v=u+at#
#v=0m/s+7/3m/s^2*12s=28m/s#

Power is given by work over time, and work is force times distance, so

#P=W/t#

#W=Fd#

#P=(Fd)/t#

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

#P=Fv#.

We can work out #F# by Newton's second law, #F=ma#,

#F=2,500,000kg*7/3m/s^2=5,833,333N#

so now,

#P=Fv#
#P=5,833,333N*28m/s=163,333,333W#

#P=163 1/3MW#