The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v = (-4.25 107) t 2 + (3.45 105) 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.)
a = m/s2
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 inside the barrel, we can start by differentiating the equation for velocity with respect to time.
Given: v = (-4.25 * 10^7) * t^2 + (3.45 * 10^5) * t
(a) Acceleration (a):
To find the acceleration, we differentiate the equation for velocity 'v' with respect to time 't'.
a = dv/dt
= d/dt [(-4.25 * 10^7) * t^2 + (3.45 * 10^5) * t]
= (-4.25 * 10^7) * 2t + (3.45 * 10^5)
= -8.5 * 10^7 * t + 3.45 * 10^5
Therefore, the acceleration of the bullet inside the barrel is a = -8.5 * 10^7 * t + 3.45 * 10^5 m/s^2.
To find the position 'x' of the bullet as a function of time 't', we integrate the equation for velocity 'v' with respect to time 't'.
x = ∫v dt
= ∫[(-4.25 * 10^7) * t^2 + (3.45 * 10^5) * t] dt
= (-4.25 * 10^7) * (t^3/3) + (3.45 * 10^5) * (t^2/2) + C
Here, 'C' is the constant of integration.
(b) Length of time the bullet is accelerated:
The bullet is accelerated while it is in the barrel. The acceleration is zero when the bullet leaves the barrel. Therefore, we need to find the time 't' at which acceleration 'a' becomes zero.
-8.5 * 10^7 * t + 3.45 * 10^5 = 0
-8.5 * 10^7 * t = -3.45 * 10^5
t = (-3.45 * 10^5) / (-8.5 * 10^7)
t ≈ 0.00406 seconds
Therefore, the bullet is accelerated for approximately 0.00406 seconds.
(c) Speed at which the bullet leaves the barrel:
To find the speed at which the bullet leaves the barrel, we substitute the time 't' at which the bullet leaves the barrel into the equation for velocity 'v'.
v = (-4.25 * 10^7) * t^2 + (3.45 * 10^5) * t
v = (-4.25 * 10^7) * (0.00406)^2 + (3.45 * 10^5) * 0.00406
v ≈ 564.556 m/s
Therefore, the speed at which the bullet leaves the barrel is approximately 564.556 m/s.
(d) Length of the barrel:
The length of the barrel can be determined by substituting the time 't' at which the bullet leaves the barrel into the equation for position 'x'.
x = (-4.25 * 10^7) * (0.00406)^3/3 + (3.45 * 10^5) * (0.00406)^2/2 + C
Since the constant of integration 'C' is not given, we cannot calculate the exact length of the barrel.