If a rocket with a mass of 3500 tons vertically accelerates at a rate of # 3/5 m/s^2#, how much power will the rocket have to exert to maintain its acceleration at 12 seconds?

Question

Charlotte Aaron · Accepted Answer

#15.12MW#

Velocity of the rocket at #12s# is

#v=u+at# #v=0m/s+3/5m/s^2*12s=7.2m/s#

Power is #"work"/"time"=W/t#, and work is #"force"*"distance"=Fd#, so

#P=W/t=(Fd)/t#

and, because velocity is distance over time, #d/t#,

#P=Fv#

#F# is found by Newtons second law,

#F=ma# #F=3,500,000kg*3/5m/s^2=2,100,000N#

so

#P=2,100,000N*7.2m/s=15,120,000Nm/s=15,120,000W#

#P=15.12MW#

Nathan Ashcroft · Answer

undefined To calculate the power the rocket must exert to maintain its acceleration, we can use the formula for power:

$$ \text{Power} = \text{Force} \times \text{Velocity} $$

The force required to maintain the acceleration can be calculated using Newton's second law of motion:

$$ \text{Force} = \text{Mass} \times \text{Acceleration} $$

Given that the mass of the rocket is 3500 tons (which is 3,500,000 kg), and the acceleration is $ \frac{3}{5} $ m/s², we can calculate the force:

$$ \text{Force} = 3,500,000 \times \frac{3}{5} $$

Now, we need to find the velocity of the rocket at 12 seconds. Assuming it starts from rest, the velocity can be calculated using the formula:

$$ \text{Velocity} = \text{Acceleration} \times \text{Time} $$

$$ \text{Velocity} = \frac{3}{5} \times 12 $$

Now, we can calculate the power:

$$ \text{Power} = \text{Force} \times \text{Velocity} $$