Question

The vertical position of a ball suspended by a rubber band is given by the equation $$ y(t)=(3.8 \mathrm{~m}) \sin (0.46 t / \mathrm{s}-0.31)-(0.2 \mathrm{~m} / \mathrm{s}) t+5.0 \mathrm{~m} $$ a) What are the equations for velocity and acceleration for this ball? b) For what times between 0 and \(30 \mathrm{~s}\) is the acceleration zero?

Short answer

Question: Find the equations for velocity (v) and acceleration (a) of an object whose position function is given by \(y(t) = (3.8 \mathrm{~m}) \sin (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s})t + 5.0 \mathrm{~m}\) in the time range 0 to 30 seconds, and find the times when the acceleration is zero. Answer: The velocity equation is given by \(v(t) = (3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s}) \cos (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s})\), and the acceleration equation is given by \(a(t) = -(3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s})^2 \sin (0.46 t / \mathrm{s} -0.31)\). The times when the acceleration is zero are approximately given by \(t_k = \frac{k\pi + 0.31}{0.46 / \mathrm{s}}\) for k = 0, 1, 2, ..., 19.

Step-by-step solution

Find the velocity equation by differentiating y(t)

Differentiating the position function y(t) with respect to time, t, gives the velocity function v(t): $$ v(t) = \frac{dy(t)}{dt} $$ Using the given equation for y(t): $$ y(t) = (3.8 \mathrm{~m}) \sin (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s})t + 5.0 \mathrm{~m} $$ Differentiate it to get the velocity equation: $$ v(t) = \frac{d}{dt}[(3.8 \mathrm{~m}) \sin (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s})t + 5.0 \mathrm{~m}] $$ The first term is a product of a constant and a sine function, while the second term is linear, and the third term is a constant. Apply the rules of differentiation for each term: $$ v(t) = (3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s}) \cos (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s}) $$

Find the acceleration equation by differentiating velocity v(t)

Differentiating the velocity function v(t) with respect to time, t, gives the acceleration function a(t): $$ a(t) = \frac{dv(t)}{dt} $$ Use the equation for v(t) found in Step 1: $$ v(t) = (3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s}) \cos (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s}) $$ Differentiate it to get the acceleration equation: $$ a(t) = \frac{d}{dt}\left[(3.8 \mathrm{~m})(0.46 / \mathrm{s}) \cos (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s})\right] $$ The first term is a product of a constant and a cosine function, while the second is a constant. Apply the rules of differentiation for each term: $$ a(t) = -(3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s})^2 \sin (0.46 t / \mathrm{s} -0.31) $$

Find the time when acceleration is zero

To find the time points when the acceleration is zero, set a(t) equal to zero and solve for t: $$ 0 = -(3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s})^2 \sin (0.46 t / \mathrm{s} -0.31) $$ We are searching for the time interval between 0 and 30 s. The sine function will be zero when its argument, \((0.46 t / \mathrm{s} - 0.31)\), is an integer multiple of \(\pi\). Thus, we can write: $$ 0.46 t / \mathrm{s} - 0.31 = k\pi $$ Solve for t: $$ t = \frac{k\pi + 0.31}{0.46 / \mathrm{s}} $$ Since we are looking for times between 0 and 30s, plug in integer values for k and check if they fall in this range. To find an approximate lower bound, plug k=0: $$ t = \frac{0\pi + 0.31}{0.46 / \mathrm{s}} \approx 0.67 \mathrm{~s} $$ To find an approximate upper bound, plug k=20: $$ t = \frac{20\pi + 0.31}{0.46 / \mathrm{s}} \approx 43.18 \mathrm{~s} $$ Since 43.18 s is outside the range, we can conclude that the valid values for k are between 0 and 20 (excluding 20). Thus, the times when the acceleration is zero are approximately: $$ t_k = \frac{k\pi + 0.31}{0.46 / \mathrm{s}} $$ for k = 0, 1, 2, ..., 19.

Key concepts

Differential Calculus in Physics

Differential calculus is a branch of mathematics that deals with the calculation of instantaneous rates of change, quantified as derivatives. In the realm of physics, differential calculus is indispensable as it enables us to derive properties such as velocity and acceleration from position-time relationships.

In our exercise, the position of a ball is described by a function involving the sine function, a linear term, and a constant. By applying differential calculus, we calculate the derivative of the position function with respect to time, which yields the velocity. Similarly, by taking the derivative of the velocity, we obtain the acceleration. This process embodies the core concept of differential calculus in physics: using mathematical tools to derive motion properties from fundamental equations of movement.

Velocity and Acceleration Equations

Understanding the equations for velocity and acceleration is vital in physics as they describe an object's state of motion. Velocity is defined as the rate of change of position with respect to time, and it is the first derivative of the position function with respect to time. Acceleration, on the other hand, is the rate of change of velocity with respect to time, thus the second derivative of the position function.

In the exercise, once we differentiate the given position equation, we obtain the velocity equation. A second differentiation gives us the acceleration equation. These equations are fundamental because they provide insight into how the position, speed, and rate of change of speed evolve over time, allowing us to predict the behavior of moving objects.

Sine Function Properties

The sine function plays a crucial role in modeling periodic phenomena such as waves and oscillations. Some key properties of the sine function that are important in our exercise include its periodicity and symmetry. The sine function is periodic with a period of \(2\pi\), meaning that it repeats values after every \(2\pi\) unit interval. Furthermore, the sine of an angle is zero whenever the angle is an integer multiple of \(\pi\).

This property is crucial when solving part b) of our exercise, where we search for times at which the acceleration—the derivative of a cosine function, which itself is derivative of a sine function—is zero. Leveraging these properties allows us to find specific intervals where a repeating pattern, such as the zero-crossing of acceleration, occurs, by solving simple trigonometric equations. As highlighted in the solution, identifying times when acceleration is zero can be done by setting the sine part of the equation to zero and solving for the time variable.