Question

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 a velocity of 200 m/s?

Answer 1

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/s​2

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  $c = 28853.78 \ m^2 /s^2$

Explanation:

From the question we are told that

         The acceleration is  $a = \frac{c}{s}\ m/s^2$

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is $s_f = 3 \ m$

          The velocity is  $v = 200 \ m/s$

     Generally  $time = \frac{ds}{dv}$

   and  acceleration is $a = \frac{v}{time }$

so

            $a = v \frac{dv}{ds}$

 =>        $vdv = a ds$

             $vdv = \frac{c}{s} ds$

integrating both sides

           $\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds$

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =$s_f$= 1.5 m

So we have  

           $\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

              $[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

            $\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

=>           $c = \frac{20000}{0.69315}$

              $c = 28853.78 \ m^2 /s^2$