The gas piston of an M4 carbine rifle with in initial position of x0=5[mm].experiences a velocity defined by x’=3+5t for one second. What is the piston’s finale position?
since x' is the velocity you need to integrate the expression 3+5t so as to get x, w/c is x = 5/2 x^2 +3t +5 where x is in mm and t in sec. Thus the final position is x(1) = 5/2 +3 +5 =21/2 mm
sorry what is the meaning of x^2
\[x^{2}\]
Sorry my dear can you write it a little simple, i didn't get it, plz step by step im writing
see, when we integrate 3+5t dt, it will be become 3t+5t square/2 + c, now how to find c
Zekarais are you there
\[x'=3+5t\]\[x =\int\limits(3+5t)dt\]\[x =\frac{ 5 }{ 2 }t^{2}+3t +c\]Since we told that the initial position is 5mm, we set t = 0 so as to get c, w/c is 5.
oka after this
Thus we will get this equation\[x =\frac{ 5 }{ 2 }t^{2}+3t+5\]
after this equation, go ahead
up to find the 21/0 ir 10.5 mm
21/2 or 10.5
To find the final position set t = 1, w/c is given.
Did you get it?
how you found the C = 5
what is w/c ?
What was the position of the piston initially?
5 mm,
so it mean C= initial position ?
5mm means the position at t = 0,right. yape
after this step, x=52t2+3t+c how to find C, or we should put the 5 directly ?
set t = 0, c= 5mm. It just a simple task.
where you gone
hmm.. please set t = 0 , c = 5 once
Dear my friend the question said " if you put t = 0 you will get the position of the piston 5mm." I think this will help
Sorry, I want to write complete step by step.
I AM OUT.
anyway thanks alot
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