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 = (-5.25 x 10^7) t^2 + (3.15 x 10^5) 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. (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.) a= [ ] m/s^2 x= [ ]m (b) Determine the length of time the bullet is accelerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel?

Explanation / Answer

a)
The acceleration is dv/dt.
v = (-5.25*10^7) t2 + (3.15*10^5) t
a= dv/dt = (-10.5*10^7)t + (3.15*10^5) m/s/s

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)(-5.25×10^7)(0)^3 + ½(3.15×10^5)(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.
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 = (-10.5*10^7)t + (3.15*10^5)

and:

x(t0) = L= (1/3)(-5.25×10^7)(t)^3 + ½(3.15×10^5)(t)²

b) a = 0 = (-10.5*10^7)t + (3.15 x 10^5)
t = 0.003 sec

c)v = (-5.25 x 10^7)t^2 + (3.15 x 10^5)t @ t=3 x 10^-3 sec
= (-5.25 *10^7)*0.003^2 + (3.15*10^5)0.003
v = 472.5 m/s

d) L =

(1/3)(-5.25×10^7)(0.003)^3 + ½(3.15×10^5)(0.003)²

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