The speed of a bullet as it travels down the barrel of a rifle toward the openin

Question

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v = (-5.50x10^7)t^2+(3.15x10^5)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.

Find the speed at which the bullet leaves the barrel.
What is the length of the barrel?

Explanation / Answer

First, arm yourself with two other equations that will help you: 1. The bullet's acceleration is just the derivative of its velocity: a = dv/dt = (-8.00×10^7)t + 2.25×10^5 2. The bullet's displacement (position in the barrel) is just the integral of its velocity: x = ?v·dt = (1/3)(-4.00×10^7)t^3 + ½(2.25×10^5)t² + C (To figure out the constant "C" in the last formula, set t = 0: x(0) = (1/3)(-4.00×10^7)(0)^3 + ½(2.25×10^5)(0)² + C x(0) = C That means "C" is the bullet's location at t=0, which we can take to be zero. So C=0.) Now that we have three formulas, we can tackle the problem: > The acceleration of the bullet becomes zero just as it reaches the end of the barrel. Call the barrel's length "L". The above sentence thus means that a = 0 at the same moment that x = L. In terms of our equations for "a" and "x", that means at the moment in time "t0" when the bullet leaves the barrel, we have: a(t0) = 0 = (-8.00×10^7)(t0) + 2.25×10^5 [Equation 1] and: x(t0) = L= (1/3)(-4.00×10^7)(t0)^3 + ½(2.25×10^5)(t0)² [Equation 2] > a) Determine the length of time the bullet is accelerated. That also means: the length of time it was in the barrel, which we've called "t0". So, solve Equation 1 above for t0: 0 = (-8.00×10^7)(t0) + 2.25×10^5 t0 = 2.25×10^5 / 8.00×10^7 > b) Find the speed at which the bullet leaves the barrel. Just plug t0 (which we've just calculated) into the equation for v: v = (-4.00×10^7)(t0)² + (2.25×10^5)(t0) > c) What is the length of the barrel? From Equation 2 above: L = (1/3)(-4.00×10^7)(t0)^3 + ½(2.25×10^5)(t0)²

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