If a rocket with a mass of #700 kg# vertically accelerates at a rate of # 3/2 m/s^2#, how much power will the rocket have to exert to maintain its acceleration at 4 seconds?

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Answer 1 · 2017-01-29T14:47:35

P=6.3 KW

see

MASS=700 kg
Acceleration=#3/2m/s^2#

THEREFORE force exerted is #m*a=700*3/2=1050N#

Now

#a=(v-u)/t# and we have U ,i.e initial velocity =0 as it started from rest and ofcourse every rockets before launching is at rest,lol.

so by this we can get #v# and #t# we are given 4 seconds

#3/2=v/4#
#v=6 m/s#

therefore

#E=F*Displacement#
dividing both sides of the equation by #t#

#E/t=F*(displacement)/t#

And by definition power is energy per unit time and velocity is displacement per unit time.so applying these

#p=F*v#
#P=1050*6# Watt

#P=6300 W#
#P= 6.3 KW#