The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v = (-4.25 10^7) t^2 + (3.45 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.
s
(c) Find the speed at which the bullet leaves the barrel.
m/s
(d) What is the length of the barrel?
m
To determine the acceleration and position of the bullet as a function of time when it is in the barrel, we need to find the derivative of the given velocity function with respect to time.
Given velocity function: v = (-4.25 * 10^7) t^2 + (3.45 * 10^5) t
(a) Acceleration (a):
To find the acceleration, we need to take the derivative of the velocity function with respect to time (t).
a = d(v)/dt
Differentiating the velocity function:
a = d/dt[(-4.25 * 10^7) t^2 + (3.45 * 10^5) t]
= (-4.25 * 10^7)(2t) + (3.45 * 10^5)
= -8.50 * 10^7 t + 3.45 * 10^5
Therefore, the acceleration of the bullet as a function of time in the barrel is:
a = -8.50 * 10^7 t + 3.45 * 10^5 (m/s^2)
To find the position of the bullet as a function of time, we need to integrate the velocity function with respect to time.
x = ∫v dt
Integrating the velocity function:
x = ∫[(-4.25 * 10^7) t^2 + (3.45 * 10^5) t] dt
= (-4.25 * 10^7) ∫t^2 dt + (3.45 * 10^5) ∫t dt
= (-4.25 * 10^7)(t^3/3) + (3.45 * 10^5)(t^2/2) + C
where C is the constant of integration.
Therefore, the position of the bullet as a function of time in the barrel is:
x = (-1.417 * 10^7) t^3 + (1.725 * 10^5) t^2 + C (m)
(b) To determine the length of time the bullet is accelerated, we need to find the time when the acceleration (a) is zero.
Setting a = 0:
-8.50 * 10^7 t + 3.45 * 10^5 = 0
t = (3.45 * 10^5) / (8.50 * 10^7)
t ≈ 0.00406 s
Therefore, the bullet is accelerated for approximately 0.00406 seconds.
(c) To find the speed at which the bullet leaves the barrel, we need to evaluate the velocity function at the time when the bullet exits the barrel (when t = 0.00406 s).
v = (-4.25 * 10^7) (0.00406)^2 + (3.45 * 10^5) (0.00406)
v ≈ -698.153 m/s
Therefore, the speed at which the bullet leaves the barrel is approximately 698.153 m/s.
(d) The length of the barrel can be found by determining the position of the bullet at the time it exits the barrel (when t = 0.00406 s).
x = (-1.417 * 10^7) (0.00406)^3 + (1.725 * 10^5) (0.00406)^2 + C
Since the position at the start of the barrel is typically assumed to be zero (x = 0) and the position at the end of the barrel is unknown, we can eliminate the constant term (C) by substituting x = 0.
0 = (-1.417 * 10^7) (0.00406)^3 + (1.725 * 10^5) (0.00406)^2 + C
Solving for C:
C ≈ 2.138 * 10^-7
Therefore, the length of the barrel is equal to the position of the bullet at the end of the barrel:
x = (-1.417 * 10^7) (0.00406)^3 + (1.725 * 10^5) (0.00406)^2 + 2.138 * 10^-7
x ≈ 8.688 * 10^-4 m
Therefore, the length of the barrel is approximately 0.0008688 meters.