The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by
v=(-5.00×10^7)t^2 + (3.00×10^5)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 functions of time when the bullet is in the barrel. (b) Determine the time inter- val over which the bullet is accelerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel?
(a) To determine the acceleration of the bullet as it travels down the barrel, we take the derivative of the velocity function with respect to time.
v(t) = (-5.00×10^7)t^2 + (3.00×10^5)t
Differentiating with respect to time, we have:
a(t) = d/dt [(-5.00×10^7)t^2 + (3.00×10^5)t]
= -10.00×10^7t + 3.00×10^5
So, the acceleration of the bullet as a function of time when it is in the barrel is given by:
a(t) = -10.00×10^7t + 3.00×10^5
To determine the position of the bullet as a function of time, we integrate the velocity function with respect to time.
s(t) = ∫ [(-5.00×10^7)t^2 + (3.00×10^5)t] dt
= (-1.67×10^7)t^3 + (1.50×10^5)t^2 + C
Where C is a constant of integration.
(b) The time interval over which the bullet is accelerated can be determined by finding the time period during which the acceleration is non-zero. The acceleration is zero when:
-10.00×10^7t + 3.00×10^5 = 0
Solving for t, we get:
t = 3.00×10^5 / 10.00×10^7
t ≈ 0.03 seconds
Therefore, the bullet is accelerated for a time interval of approximately 0.03 seconds.
(c) The speed at which the bullet leaves the barrel is given by the velocity function when t = 0. To find this, we substitute t = 0 into the velocity function:
v(0) = (-5.00×10^7)(0)^2 + (3.00×10^5)(0)
= 0
Thus, the speed of the bullet as it leaves the barrel is 0 m/s.
(d) The length of the barrel can be determined by finding the distance travelled by the bullet when it reaches the end of the barrel. This occurs when t equals the time it takes for the bullet to exit the barrel. To find this, we set the position function equal to the length of the barrel:
(-1.67×10^7)t^3 + (1.50×10^5)t^2 + C = L
Where L is the length of the barrel and C is the constant of integration.
Since the bullet is at rest initially (t = 0), we can eliminate the cubic term of the position equation, leaving us with:
(1.50×10^5)t^2 + C = L
Substituting t = 0.03 seconds (from part b) and v = 0 m/s (from part c), we can solve for C:
(1.50×10^5)(0.03)^2 + C = 0
C = -1.35
Now we can find the length of the barrel by setting t equal to the time it takes for the bullet to exit the barrel and solving for L:
(1.50×10^5)t^2 - 1.35 = L
Since t is the time it takes for the bullet to exit the barrel, we need to solve for t when v = 0:
0 = (-5.00×10^7)t^2 + (3.00×10^5)t
0 = t[(-5.00×10^7)t + (3.00×10^5)]
We have t = 0 as one solution, but this corresponds to the initial time (t = 0) when the bullet is in the barrel. Therefore, we discard this solution.
Setting the bracketed expression equal to zero, we can solve for the other value of t:
(-5.00×10^7)t + (3.00×10^5) = 0
t ≈ 0.006 seconds
Now, substituting t = 0.006 seconds into the position function, we get:
(1.50×10^5)(0.006)^2 - 1.35 = L
L ≈ 5.4 meters
Therefore, the length of the barrel is approximately 5.4 meters.
(a) To determine the acceleration and position of the bullet as functions of time when the bullet is in the barrel, we need to find the derivatives of the velocity function.
Given:
v = (-5.00×10^7)t^2 + (3.00×10^5)t
Acceleration (a) is the derivative of velocity (v) with respect to time (t).
a = dv/dt
Taking the derivative of the velocity function, we get:
a = d/dt[(-5.00×10^7)t^2 + (3.00×10^5)t]
a = -10.00×10^7t + 3.00×10^5
The position (x) as a function of time can be found by integrating the velocity function. We'll also include an integration constant (C) since we are dealing with position.
x = ∫v dt + C
Integrating the velocity function, we get:
x = ∫((-5.00×10^7)t^2 + (3.00×10^5)t) dt
x = (-1.67×10^7)t^3 + (1.50×10^5)t^2 + C
(b) To determine the time interval over which the bullet is accelerated, we need to find the values of time when the acceleration is zero.
Setting the acceleration equation equal to zero and solving for t:
-10.00×10^7t + 3.00×10^5 = 0
t = 3.00×10^5 / (10.00×10^7)
t = 0.03 seconds
So, the bullet is accelerated for a time interval of 0 to 0.03 seconds.
(c) To find the speed at which the bullet leaves the barrel, we can substitute the time when the bullet leaves the barrel into the velocity equation.
Substituting t = 0.03 seconds into the velocity equation:
v = (-5.00×10^7)(0.03)^2 + (3.00×10^5)(0.03)
v = -45,000 + 9,000
v = -36,000 m/s
The speed at which the bullet leaves the barrel is 36,000 m/s.
(d) The length of the barrel can be found by calculating the position at the time the bullet leaves the barrel.
Substituting t = 0.03 seconds into the position equation:
x = (-1.67×10^7)(0.03)^3 + (1.50×10^5)(0.03)^2 + C
x = -15.03 meters + 135 meters + C
x = 119.97 + C meters
Since C is just an integration constant, it represents the initial position of the bullet in the barrel. Assuming the bullet starts at the beginning of the barrel, C = 0.
Therefore, the length of the barrel is approximately 119.97 meters.
I will do position last
a = dv/dt
so
a = -10*10^7 t + 3*10^5
zero at
t = 3/10 * 10^-2 = .003 seconds
a is positive for the whole .003s
find v at t = .003 for part c
x = integral v dt
with x = 0 at t = 0
so
x = (1/3)(-5.00×10^7)t^3 +(1/2)(3.00×10^5)t^2
length of barrel is x when t = .003