SOLUTION: A 9800 kg rocket is traveling east along a horizontal frictionless rail at a velocity of 11 m/s. The rocket is then accelerated uniformly to a velocity of 22 m/s in a time of 0.75

Question 1100855: A 9800 kg rocket is traveling east along a horizontal frictionless rail at a velocity of 11 m/s. The rocket is then accelerated uniformly to a velocity of 22 m/s in a time of 0.75 s by the expulsion of hot gases. What is the average force with which the gases are expelled by the rocket?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

One of the kinematics equations is

a = (Vf-Vi)/t

where,

a = acceleration
Vi = initial velocity
Vf = final velocity
t = elapsed time

In this case,
a = unknown (we'll find it shortly)
Vi = 11 meters per second
Vf = 22 meters per second
t = 0.75 seconds

Plug the three given values into the equation and compute
a = (Vf-Vi)/t
a = (22-11)/0.75
a = 14.667 (accurate to 3 decimal places)

The rocket's acceleration in this time frame is roughly 14.667 meters per second per second. Basically each second, the velocity is increasing by roughly 14.667 m/s.

The mass is given to be 9800 kg, so m = 9800.

Use both of these values to find the force F. Specifically, we will use Newton's 2nd Law that says F = m*a (force = mass times acceleration)

So,
F = m*a
F = 9800*14.667
F = 143,736.6

The gasses are expelled with an average force of roughly 143,736.6 Newtons; which accelerates the rocket from 11 m/s to 22 m/s in a timespan of 0.75 seconds.

Note: if you need a more accurate answer, then use a more accurate value for the acceleration.

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