Problem 55 - University Calculus: Early Transcendentals

Question

The equations give the position \(s=f(t)\) of a body moving on a coordinate line ( \(s\) in meters, \(t\) in seconds). Find the body's velocity, speed, acceleration, and jerk at time \(t=\pi / 4 \mathrm{sec}\). $$s=2-2 \sin t$$

Step-by-Step Solution

Answer

Velocity: \(-\sqrt{2}\) m/s, Speed: \(\sqrt{2}\) m/s, Acceleration: \(\sqrt{2}\) m/s², Jerk: \(\sqrt{2}\) m/s³.

Step 1: Find the velocity function

To find the velocity of the body, we need to take the derivative of the position function \(s(t) = 2 - 2 \sin(t)\) with respect to time \(t\). Recall that the derivative of \(\sin(t)\) is \(\cos(t)\). Thus, the velocity function \(v(t)\) is:\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(2 - 2 \sin(t)) = -2 \cos(t) \]

Step 2: Calculate the velocity at \(t = \pi/4\)

Substitute \(t = \pi/4\) into the velocity function found in Step 1:\[ v(\pi/4) = -2 \cos(\pi/4) = -2 \left( \frac{\sqrt{2}}{2} \right) = -\sqrt{2} \, \text{m/s} \]

Step 3: Find the speed at \(t = \pi/4\)

Speed is the absolute value of velocity. At \(t = \pi/4\), the velocity is \(-\sqrt{2} \, \text{m/s}\). Thus, speed is:\[ |v(\pi/4)| = | -\sqrt{2} | = \sqrt{2} \, \text{m/s} \]

Step 4: Find the acceleration function

Acceleration is the derivative of velocity. Differentiate the velocity function \(v(t) = -2 \cos(t)\) with respect to \(t\). The derivative of \(\cos(t)\) is \(-\sin(t)\), thus:\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(-2 \cos(t)) = 2 \sin(t) \]

Step 5: Calculate the acceleration at \(t = \pi/4\)

Substitute \(t = \pi/4\) into the acceleration function:\[ a(\pi/4) = 2 \sin(\pi/4) = 2 \left( \frac{\sqrt{2}}{2} \right) = \sqrt{2} \, \text{m/s}^2 \]

Step 6: Find the jerk function

Jerk is the derivative of acceleration. Differentiate the acceleration function \(a(t) = 2 \sin(t)\) with respect to \(t\). The derivative of \(\sin(t)\) is \(\cos(t)\), thus:\[ j(t) = \frac{da}{dt} = \frac{d}{dt}(2 \sin(t)) = 2 \cos(t) \]

Step 7: Calculate the jerk at \(t = \pi/4\)

Substitute \(t = \pi/4\) into the jerk function:\[ j(\pi/4) = 2 \cos(\pi/4) = 2 \left( \frac{\sqrt{2}}{2} \right) = \sqrt{2} \, \text{m/s}^3 \]

Key Concepts

Understanding Velocity in Calculus

Velocity is a fundamental concept in calculus and physics, representing the rate of change of position with respect to time. When we differentiate a position function, we obtain the velocity function. For example, in the given exercise, the position function is expressed as \(s(t) = 2 - 2 \sin(t)\). By calculating the derivative of this function with respect to \(t\), we find the velocity function: \(v(t) = -2 \cos(t)\).

Finding the velocity at a specific time involves substituting the time value into the velocity function. At \(t = \pi/4\), the velocity is calculated as \(v(\pi/4) = -\sqrt{2} \text{ m/s}\), indicating the speed and direction of the movement at that moment.

• Velocity is directional and can be negative or positive.
• Speed is the magnitude of velocity and is always positive, calculated as the absolute value of velocity.

By understanding velocity, we can interpret how and where an object moves within a given timeframe.

Acceleration: The Change in Velocity

Acceleration represents the rate of change of velocity over time. It tells us how quickly an object is speeding up or slowing down. To find the acceleration, we take the derivative of the velocity function. From the exercise, we have the velocity function \(v(t) = -2 \cos(t)\).

Differentiating this function gives us the acceleration function \(a(t) = 2 \sin(t)\). This shows how velocity is changing at every moment in time.

Calculating acceleration at \(t = \pi/4\), we see \(a(\pi/4) = \sqrt{2} \text{ m/s}^2\), meaning that the object is accelerating at this rate at that time.

• Acceleration can be positive (speeding up) or negative (slowing down), also known as deceleration.
• Acceleration provides insight into the dynamics of the object's movement.

Understanding acceleration is crucial for predicting how an object's velocity will change over time.

Exploring Jerk in Motion

Jerk is the derivative of acceleration, indicating how acceleration changes over time. Unlike velocity and acceleration, jerk is often less intuitive but is essential in scenarios where the change in acceleration is non-negligible, such as in mechanical engineering or designing systems for comfort.

From our exercise, the acceleration function is \(a(t) = 2 \sin(t)\), and differentiating it gives the jerk function: \(j(t) = 2 \cos(t)\). This new function tells us how the acceleration itself is increasing or decreasing.

Calculating the jerk at \(t = \pi/4\), we find \(j(\pi/4) = \sqrt{2} \text{ m/s}^3\).

• Jerk is particularly relevant in the analysis of systems requiring smooth or sudden transitions.
• Understanding jerk can improve design and performance in engineering solutions.

In essence, jerk helps refine our understanding of motion by considering the change in acceleration itself, adding a deeper layer to motion analysis.