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?

Question

1744 views around the world

You can reuse this answer
Creative Commons License

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#