Question

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?

Answer 1

P=6.3 KW

Explanation:

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`