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 = (4.80 107)t2 + (2.45 105)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.

(a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel. (Use t as necessary and round all numerical coefficients to exactly 3 significant figures.)

(b) Determine the length of time the bullet is accelerated.
s

(c) Find the speed at which the bullet leaves the barrel.
m/s

(d) What is the length of the barrel?
m

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 )acceleration = -8.00*10^7t + 2.45 * 10^5

position of bullet with as a function of time = (-1.6*10^7)t^3 +( 1.2*10^5)t^2

> b) 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) + 2.45×10^5

t0 = 2.45×10^5 / 8.00×10^7 =3.06*10^-3

> c) 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 + (2.45×10^5)(t0)

= (-4.00×10^7)(3.06*10^-3)^2 + (2.45×10^5)(3.06*10^-3) = 375m/s

> d) What is the length of the barrel?

From Equation 2 above:

(1/3)(-4.00×10^7)(t0)^3 + ½(2.25×10^5)(t0)^2

L = (1/3)(-4.00×10^7)((3.06*10^-3)^3 + ½(2.45×10^5)((3.06*10^-3)^2 = 0.765m

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