The vertical position of a ball suspended by a rubber band is given by the equation
y(t) = (3.4 m)sin(0.49 t/s − 0.40) − (0.4 m/s)t + 5.4 m.
(a) What are the equations for velocity and acceleration for this ball? (Use the following as necessary: t.)
(b) For what times between 0 and 30 s is the acceleration zero? (Enter your answers as a comma-separated list.)
(a) Well, let's do some math acrobatics here! To find the velocity function, we need to take the derivative of the position function, y(t). And when I say derivative, I'm not talking about a creepy dating app for mathematicians! So, grab your calculators and let's get derivative!
The derivative of sin(t) is cos(t), and the derivative of t is 1. So, the velocity function, v(t), is
v(t) = (3.4 m)cos(0.49 t/s − 0.40) − (0.4 m/s).
Now, for the acceleration function, we gotta get derivative-happy once again! Taking the derivative of cos(t) gives us -sin(t). So, the acceleration function, a(t), is
a(t) = -(3.4 m)sin(0.49 t/s − 0.40).
(b) Alrighty, let's find when the acceleration is zero between 0 and 30 seconds! Remember, we don't want any zeroes trying to crash the acceleration party! So, we set a(t) equal to zero and solve for t:
-(3.4 m)sin(0.49 t/s − 0.40) = 0.
Now, for sin(t) to be zero, t must be an integer multiple of π. So, we have:
0.49 t/s - 0.40 = nπ, where n is an integer.
Solving for t:
0.49 t/s = nπ + 0.40,
t = (nπ + 0.40) / 0.49.
Now, let's find the values of t between 0 and 30 seconds that satisfy this equation. Just plug in different integers for n and check for values within the given bounds. Be careful, this is not a clowning matter!
To determine the equations for velocity and acceleration, we need to take the derivatives of the position equation with respect to time.
(a) Velocity equation:
The derivative of the position equation, y(t), with respect to time, t, gives us the equation for velocity, v(t). So, let's calculate the derivative:
y(t) = (3.4 m)sin(0.49 t/s − 0.40) − (0.4 m/s)t + 5.4 m
dy/dt = d/dt [ (3.4 m)sin(0.49 t/s − 0.40) − (0.4 m/s)t + 5.4 m ]
To differentiate the terms separately:
d/dt [ (3.4 m)sin(0.49 t/s − 0.40) ] = (3.4 m) * d/dt [sin(0.49 t/s − 0.40)]
d/dt [ (0.4 m/s)t ] = (0.4 m/s) * d/dt[t] = 0.4 m/s
d/dt [ 5.4 m ] = 0
Using the chain rule, d/dt [sin(0.49 t/s − 0.40)] can be calculated:
d/dt [sin(u)] = cos(u) * d/dt[u]
Let u = 0.49 t/s − 0.40
d/dt [sin(0.49 t/s − 0.40)] = cos(0.49 t/s − 0.40) * d/dt[0.49 t/s − 0.40]
d/dt[0.49 t/s − 0.40] = 0.49/s * d/dt[t]
Substituting back into the equation:
dy/dt = (3.4 m) * cos(0.49 t/s − 0.40) * 0.49/s - 0.4 m/s
Thus, the equation for velocity, v(t), is:
v(t) = (3.4 m) * cos(0.49 t/s − 0.40) * 0.49/s - 0.4 m/s
Acceleration equation:
Now, we need to differentiate the velocity equation with respect to time to find the acceleration equation, a(t).
dv/dt = d/dt [(3.4 m) * cos(0.49 t/s − 0.40) * 0.49/s - 0.4 m/s]
Using the chain rule, we differentiate cos(0.49 t/s − 0.40):
d/dt [cos(u)] = -sin(u) * d/dt[u]
Let u = 0.49 t/s − 0.40
d/dt [cos(0.49 t/s − 0.40)] = -sin(0.49 t/s − 0.40) * d/dt[0.49 t/s − 0.40]
d/dt[0.49 t/s − 0.40] = 0.49/s * d/dt[t]
Substituting back into the equation:
dv/dt = (3.4 m) * (-sin(0.49 t/s − 0.40)) * 0.49/s
Simplifying further:
dv/dt = -1.6666 m/s^2 * sin(0.49 t/s − 0.40)
Thus, the equation for acceleration, a(t), is:
a(t) = -1.6666 m/s^2 * sin(0.49 t/s − 0.40)
(b) To find the times between 0 and 30 seconds where the acceleration is zero, we set the acceleration equation equal to zero and solve for t:
-1.6666 m/s^2 * sin(0.49 t/s − 0.40) = 0
sin(0.49 t/s − 0.40) = 0
To solve for t, we find the values of t that make sin(0.49 t/s − 0.40) equal to zero. These values correspond to the times when the acceleration is zero.
Sin(x) = 0 whenever x is an integral multiple of π (pi). Therefore, we need to solve:
0.49 t/s – 0.40 = nπ, where n is an integer.
Simplifying the equation, we solve for t:
0.49 t/s = nπ + 0.40
t = (nπ + 0.40) / 0.49
Now, we substitute different values for n (including negative and positive integers) in the equation to find the corresponding times t between 0 and 30 seconds when the acceleration is zero. We list all the values found, separated by commas.