The speed of a bullet.............?

Question

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by the expression

v = (-4.00 x 10^7)t^2 + (2.25 x 10^5)t,

where v is in meters per second and t is in seconds. The acceleration of the bullet becomes zero just as it reaches the end of the barrel.

a) Determine the length of time the bullet is accelerated.

b) Find the speed at which the bullet leaves the barrel.

c) What is the length of the barrel?

I don't know where to start! Any help is much appreciated.

Answer 1

Without having a physics book readily available, this is basically off the top of my head:

A) Usings Newton's kinematics equations, solve for t by isolating the variable on the left side of the equation.

B) Remember that speed is the absolute value of the velocity, i.e. plug in the value obtained in A for v and solve.

C) Think of the barell as a cartesian co-ordinate system where its lenght is in the positive X plane and the bullet starting at 0. Use the kinematics equation solving for X (position) by plugging in the values obtained in A and B.

Like i said this is off the top of my head and it has been a couple years since my physics classes so verify elsewhere. Hope it points you in the right direction.

Answer 2

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

Answer 3

You have the eq for v;

a = dv/dt = -8E7*t + 2.25E5

for a = 0, 0 = -8E7*t + 2.25E5 → t = .0028125 sec

b)  v = -4E7*.0028125² + 2.25E5*.0028125 = 316.4 m/s

c) L = ∫vdt = ∫(-4E7t² + 2.25E5t)dt (from 0 to t = .0028125)

L = .593 m