The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by the expression v = (-5.15 multipli
ed by 107) t2 + (2.30 multiplied by 105) t, where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero. (b) Determine the length of time the bullet is accelerated.
(c) Find the speed at which the bullet leaves the barrel.
I'm assuming you're in calculus based physics. I apologize if otherwise.
Let's come up with equations for distance and acceleration as functions of time.
From the definition of acceleration we know that<span>a=<span><span>dv</span><span>dt</span></span></span>Taking the derivative of v with respect to t yields<span>a=<span><span>dv</span><span>dt</span></span>=(−5.15∗107)∗2∗t+(2.30∗105)</span>
From the definition of distance, we know that <span>v=<span><span>dx</span><span>dt</span></span>→dx=vdt</span>Integrating velocity yields<span>x=(−5.15∗107)∗<span>(<span><span>t3</span>3</span>)</span>+(2.30∗105)∗<span>(<span><span>t2</span>2</span>)</span>+<span>x0</span></span> where <span>x0</span> is the starting position.
If acceleration is zero when the bullet leaves the barrel, we can use our equation for acceleration to determine the time the bullet is in the barrel. This is seen as<span>0=(−5.15∗107)∗2∗t+(2.30∗105)→t=<span><span>−(2.30∗105)</span><span>(−5.15∗107)</span></span></span>
Knowing time, we can solve for the velocity as the bullet leaves the barrel by plugging time back into our given equation for velocity.
The length of the barrel can be solved by plugging time back into our equation for distance and setting <span><span>x0</span>=0</span><span>.</span>
