# 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?

Jul 8, 2018

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,

${F}_{\text{net}} = m a$

where:

• ${F}_{\text{net}}$ 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.

$\therefore {F}_{\text{net}} = m a + m g$

$= m \left(a + g\right)$

Substituting our values, we get:

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

$= 41632500 \setminus \text{N}$

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

$v = u + a t$

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:

$v = 0 + 0.875 \setminus \text{m/s"^2*5 \ "s}$

$= 4.375 \setminus \text{m/s}$

Power is given by the equation:

$P = F \cdot v$

where:

• $P$ is the power in watts

• $F$ is the force in newtons

• $v$ is the velocity in meters per second

So, we get:

$P = 41632500 \setminus \text{N"*4.375 \ "m/s}$

$= 182142188 \setminus \text{W}$

$\approx 182.14 \setminus \text{MW}$