Question

If a rocket with a mass of 3900 tons vertically accelerates at a rate of `7/8 m/s^2`, how much power will the rocket have to exert to maintain its acceleration at 5 seconds?

Answer 1

Approximately `182.14` megawatts.

Explanation:

We first find the net force acting on the rocket using Newton's second law of motion, which states that,

`F_"net"=ma`

where:

• `F_"net"` is the net force in newtons

• `m` is the mass in kilograms

• `a` is the acceleration in meters per second squared

Here, the net force is the force applied to the rocket by acceleration plus the rocket's weight.

`:.F_"net"=ma+mg`

`=m(a+g)`

Substituting our values, we get:

`F_"net"=3900000 \ "kg"(0.875 \ "m/s"^2+9.8 \ "m/s"^2)`

`=41632500 \ "N"`

Now, we find the velocity of the rocket, given by the equation:

`v=u+at`

where:

• `v` is the final velocity

• `u` is the initial velocity

• `a` is the acceleration

• `t` is the time taken

Assuming `u=0`, we find the rocket's speed after five seconds:

`v=0+0.875 \ "m/s"^2*5 \ "s"`

`=4.375 \ "m/s"`

Power is given by the equation:

`P=F*v`

where:

• `P` is the power in watts

• `F` is the force in newtons

• `v` is the velocity in meters per second

So, we get:

`P=41632500 \ "N"*4.375 \ "m/s"`

`=182142188 \ "W"`

`~~182.14 \ "MW"`