Maximum and average power for a 120 kg shell fired from a 37.5 m barrel

[cheff3r]

I state my problen in my solution
A big gun had a range of 120 km, a 37.5 m barrel, a muzzle velocity of 1550 m/s and fired a 120 kg shell.
a) If you assume the shell’s acceleration to be constant, what maximum power was transferred to the shell as it traveled along the barrel? [5 Marks]
b) What average power was transferred to the shell? [3 Marks


MY solution

okay so i went about it finding acceleration
u=0
v=1550
s=37.5
m=120

v^2=u^2+2as
a=(v^2-u^2)/(2s)

so now i have a=32033.33m/s^2

then using Newtons law

F=ma
F=120*32033.33

then

F=3844000N

W=Fd
this is my first question is it fair to say d=37.5m or is it 120km??
well that's what i have done

W=144150000J

P=w/t
second question how do i find time?

third question is the above finds the total power yes? how do i find max power, does it mean its the power at a point? and is average pwoer just total power/distance? to find part b

 

[Kurdt]

For part a you would use the length of the barrel. The other info given is that the shell undergoes constant acceleration in the barrel and has a final speed of 1550m/s at the end of the barrel. One can work out the time from that. Maximum power I think will be instantaneous power, which is the force multiplied by the velocity. Average power is what you've been calculating.

 

[cheff3r]

For part a you would use the length of the barrel. The other info given is that the shell undergoes constant acceleration in the barrel and has a final speed of 1550m/s at the end of the barrel. One can work out the time from that. Maximum power I think will be instantaneous power, which is the force multiplied by the velocity. Average power is what you've been calculating.

ahh so you mean the v=u+at and solve for t

and that makes sense about the max power thanks

 

[Kurdt]

ahh so you mean the v=u+at and solve for t

Yes, that is what I was getting at.

 

[actionman26]

This looks familiar "yes or no?"