Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is [tex]c = 28853.78 \ m^2 /s^2[/tex]
Explanation:
From the question we are told that
The acceleration is [tex]a = \frac{c}{s}\ m/s^2[/tex]
The initial position of the projectile is s= 1.5m
The final position of the projectile is [tex]s_f = 3 \ m[/tex]
The velocity is [tex]v = 200 \ m/s[/tex]
Generally [tex]time = \frac{ds}{dv}[/tex]
and acceleration is [tex]a = \frac{v}{time }[/tex]
so
[tex]a = v \frac{dv}{ds}[/tex]
=> [tex]vdv = a ds[/tex]
[tex]vdv = \frac{c}{s} ds[/tex]
integrating both sides
[tex]\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds[/tex]
Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d =[tex]s_f[/tex]= 1.5 m
So we have
[tex]\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds[/tex]
[tex][\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.[/tex]
[tex]\frac{200^2}{2} = c ln[\frac{3}{1.5} ][/tex]
=> [tex]c = \frac{20000}{0.69315}[/tex]
[tex]c = 28853.78 \ m^2 /s^2[/tex]
