If a rocket with a mass of 4000 tons vertically accelerates at a rate of 9/5 m/s^2, how much power will the rocket have to exert to maintain its acceleration at 11 seconds?

1 Answer

P=8,74*10^9W=8,74GW

Explanation:

The power at a single moment is given by the formula :

P=vF

Let's find the force first :

ΣF=ma=>F-mg=ma=>F=m(g+a)=

4*10^6(9,81+4,5)=176,6*10^6=1.766*10^8N

Now let's find the velocity :

v=at=4,5*11=49,5m/s

The power is therefore :

P=vF=49,5*1,766*10^8=87,4*10^8=8,74*10^9W=

8,74GW