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 the velocity and acceleration of this ball? b) For what times between 0 and 30 s is the acceleration zero?

Short answer

a) The velocity equation for the ball is: $$ v(t) = 1.748 \cos(0.46t - 0.31) - 0.2 $$ and the acceleration equation is: $$ a(t) = -0.80308 \sin(0.46t - 0.31) $$ b) The times between 0 and 30 seconds when the acceleration is zero are given by: $$ t = (-\pi + 0.31)/0.46, \quad(\pi + 0.31)/0.46, \quad(3\pi + 0.31)/0.46, \quad \ldots, \quad(20\pi + 0.31)/0.46 $$

Step-by-step solution

Finding the velocity equation

First, let's find the derivative of the position function with respect to time, which will give us the velocity equation. The position function is given by: $$ y(t) = (3.8 \mathrm{~m})\sin(0.46t - 0.31) - (0.2 \mathrm{~m/s})t + 5.0 \mathrm{~m} $$ Taking the derivative with respect to time, we get: $$ v(t) = \frac{dy(t)}{dt} = (3.8) \cdot (0.46) \cos(0.46t - 0.31)-0.2 $$ So, the velocity equation is: $$ v(t) = 1.748 \cos(0.46t - 0.31) - 0.2 $$

Finding the acceleration equation

Now, let's find the derivative of the velocity function with respect to time, which will give us the acceleration equation. Taking the derivative of the velocity function we found in Step 1, we get: $$ a(t) = \frac{dv(t)}{dt} = -1.748 \cdot (0.46) \sin(0.46t - 0.31) $$ So, the acceleration equation is: $$ a(t) = -0.80308 \sin(0.46t - 0.31) $$

Finding the times when acceleration is zero

Now, we need to find the times when the acceleration is zero between 0 and 30 seconds. We will solve the equation $$a(t) = 0$$ for t within the given interval. From the acceleration equation, we have: $$ -0.80308 \sin(0.46t - 0.31) = 0 $$ and therefore $$ \sin(0.46t - 0.31) = 0 $$ Let's denote $$x = 0.46t - 0.31$$ and solve for $$x$$: $$ \sin(x) = 0 $$ This equation has solutions for $$x = n\pi$$, where $$n$$ is an integer. Now, we will solve for $$t$$ using the calculated values of $$x$$: $$ 0.46t - 0.31 = n\pi $$ $$ t = (n\pi + 0.31) / 0.46 $$ We only need to consider the values of $$n$$ that will give us a time between 0 and 30 seconds, which are $$n = -1, 0, 1, 2, 3,\ldots, 20$$. We can manually verify that the times from $$n=-1$$ to $$n=20$$ will be in the required range. Finally, we get the following times when the acceleration is zero: $$ t = (-\pi + 0.31)/0.46, \quad(\pi + 0.31)/0.46, \quad(3\pi + 0.31)/0.46, \quad \ldots, \quad(20\pi + 0.31)/0.46 $$

Key concepts

Derivative of Position Function

Understanding the derivative of the position function is essential in physics, especially when examining objects in motion. In simple terms, it involves calculating how the position of an object changes over time, which in turn describes its velocity. In the provided exercise, the position of a ball suspended by a rubber band is described by a sinusoidal function of time.

Let's dissect this concept a bit more: the position function given is in the form of a sine wave combined with a linear term. To comprehend the ball's movement, it's crucial to calculate the instantaneous velocity at any time 't'. To do this, we differentiate the position function with respect to time using the rules of calculus.

When we talk about 'differentiating', we're essentially looking at the slope of the position function at every point in time. If you visualize the function as a curve on a graph, the derivative at any point is the tangent to the curve at that point. The steepness of the tangent represents the velocity—the steeper it is, the faster the object is moving at that moment. This calculation steps us directly into the realm of the object's velocity equation of motion.

Velocity Equation of Motion

Once we've found the velocity equation through differentiation of the position function, it opens up a deeper understanding of the object's motion. The velocity function, as derived from the exercise, gives us a formula that we can use to find the speed and direction of the object at any given time.

The velocity equation itself is the result of applying the derivative to the position function. This means it tells us not just how fast the ball is moving, but also in what direction. In this case, since the derivative includes a cosine function, we know the velocity will follow a harmonic pattern as well, oscillating back and forth over time.

Additionally, the equation shows that the ball has a constant velocity component, indicated by the subtraction of 0.2 m/s. This represents a steady rate of change in the downward direction, likely due to the effect of the rubber band or gravity. By setting the velocity to zero, we could also find the time when the ball momentarily stops before reversing its direction, which is a critical point called the 'turning point' in motion studies.

Acceleration Equation of Motion

Understanding the acceleration of an object is integral to the analysis of its motion, and it's especially important when that motion is oscillatory, as in this exercise. Acceleration can be thought of as the rate of change of velocity, and it's obtained by differentiating the velocity function with respect to time.

The acceleration equation extracted from the exercise tells us how quickly and in which direction the object's velocity changes. A sine function in the acceleration equation indicates that the acceleration, like the position and velocity, is also harmonic and periodically changes its sign from positive to negative and vice versa, aligning with the oscillatory motion of the ball.

Importantly, the acceleration is zero when the velocity function reaches its maximum and minimum values—this happens when the ball is at the peak of its upward or downward motion, just as it starts to reverse direction. Therefore, determining when the acceleration is zero, as calculated in the exercise, helps us predict when the ball will change its direction. It is during these moments of zero acceleration that the ball has no change in velocity; it's moving neither faster nor slower for an instant, which is a fascinating point of study in harmonic motion.