The acceleration of a particle is defined by the relation a = -k/x . It has been experimentally determined that v = 15 ft/s when x = 0.6 ft and that v = 9 ft/s when x = 1.2 ft. determine (a) the velocity of the particle when x = 1.5 ft and (b) the position of the particle at which its velocity is zero.
v = -k ln x + c
15 = -k ln .6 + c
9 = -k ln 1.2+ c
-----------------subtract
6 = -k ln(.6/1.2)
etc
To solve this problem, we will use the given information and the relationship between acceleration, velocity, and position.
Given:
Acceleration, a = -k/x
Velocity, v = 15 ft/s at x = 0.6 ft
Velocity, v = 9 ft/s at x = 1.2 ft
Step 1: Determining the value of k
We can find the value of k by substituting the given values for velocity and position into the acceleration equation.
At x = 0.6 ft:
a = -k/x
-15 = -k/0.6
k = 15 * 0.6
k = 9
Therefore, k = 9.
Step 2: Calculating the velocity when x = 1.5 ft (part a)
Now that we have the value of k, we can use it to find the acceleration at x = 1.5 ft.
At x = 1.5 ft:
a = -k/x
a = -9/1.5
a = -6 ft/s^2
We now have the acceleration at x = 1.5 ft. To find the velocity, we need to integrate the acceleration with respect to time.
Step 3: Integrating acceleration to find velocity
Given a = dv/dt, we can integrate the acceleration over time to find velocity.
∫ dv = ∫ -6 dt
v = -6t + C
To find the constant C, we will use the given information for x = 0.6 ft, v = 15 ft/s:
15 = -6(0.6) + C
C = 18.6
Therefore, the equation for velocity is v = -6t + 18.6.
Step 4: Calculating the velocity for x = 1.5 ft
Substituting x = 1.5 ft into the velocity equation:
v = -6(1.5) + 18.6
v = -9 + 18.6
v = 9.6 ft/s
Therefore, the velocity of the particle when x = 1.5 ft is 9.6 ft/s.
Step 5: Finding the position where velocity is zero (part b)
To find the position where the velocity is zero, we need to solve the velocity equation:
v = -6t + 18.6
0 = -6t + 18.6
6t = 18.6
t = 3.1 s
Since velocity is zero at t = 3.1 s, we can find the position by substituting this value into the position equation.
x = -0.5kt^2 + Ct + C'
Given k = 9 and C = 18.6:
x = -0.5(9)(3.1)^2 + 18.6(3.1) + C'
x = -42.97 + 57.66 + C'
x = 14.69 + C'
Therefore, the position of the particle where its velocity is zero is at x = 14.69 + C'.
To solve the problem, we need to find a function for velocity v in terms of x. Let's integrate the given acceleration function with respect to time to find the velocity function.
Given: a = -k/x
Integrating both sides with respect to time yields:
∫a dt = -k ∫1/x dx
Using the given information, we can determine the constant k. When x = 0.6 ft, v = 15 ft/s:
15 = -k/0.6
Solving for k:
k = -9 ft/s^2
The acceleration function becomes:
a = 9/x
Now, we can integrate a to get the velocity function with respect to x:
∫dv = ∫9/x dx
ln|v| = 9 ln|x| + C
Taking the exponential of both sides:
|v| = e^(9 ln|x| + C)
Since v must be positive, we can drop the absolute value sign:
v = e^(9 ln|x| + C)
Using properties of logarithms and exponents, we can simplify the equation further:
v = e^(ln|x|^9 + C)
v = x^9 * e^C
Now, let's use the given information when x = 1.2 ft, v = 9 ft/s:
9 = 1.2^9 * e^C
Solving for C:
C = ln(9 / 1.2^9)
Substituting the value of C back into the velocity equation:
v = x^9 * e^(ln(9 / 1.2^9))
v = x^9 * (9 / 1.2^9)
(a) To find the velocity when x = 1.5 ft:
v = 1.5^9 * (9 / 1.2^9)
v ≈ 4.574 ft/s
Therefore, the velocity of the particle when x = 1.5 ft is approximately 4.574 ft/s.
(b) To find the position when the velocity is zero, we need to find the value of x that makes v zero:
0 = x^9 * (9 / 1.2^9)
We can see that for v to be zero, x must equal zero. However, this value does not satisfy the original equation a = -k/x. Therefore, there is no position at which the velocity is zero.