Problem 46 - Thomas Calculus

Question

The equations in Exercises 45 and 46 give the position \(s=f(t)\) of a body moving on a coordinate line \((s \text { in meters, } t \text { in seconds). Find the }\) body's velocity, speed, acceleration, and jerk at time \(t=\pi / 4\) sec. $$ s=\sin t+\cos t $$

Step-by-Step Solution

Answer

Velocity: 0, Speed: 0, Acceleration: \(-\sqrt{2}\), Jerk: 0.

Step 1: Find the Velocity

To find the velocity, we need to differentiate the position function \(s=f(t)\) with respect to time \(t\). Given \(s(t) = \sin t + \cos t\), the velocity \(v(t)\) is \(v(t) = \frac{ds}{dt} = \cos t - \sin t\). Now, substitute \(t=\frac{\pi}{4}\): \[ v\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0 \].

Step 2: Calculate Speed

Speed is the absolute value of velocity. Since we found that the velocity at \(t=\frac{\pi}{4}\) is 0, the speed is \[ |v\left(\frac{\pi}{4}\right)| = |0| = 0 \].

Step 3: Find the Acceleration

Acceleration is the derivative of velocity with respect to time. Using \(v(t) = \cos t - \sin t\), the acceleration \(a(t)\) is \(\frac{dv}{dt} = -\sin t - \cos t\). Substitute \(t=\frac{\pi}{4}\): \[ a\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} \].

Step 4: Determine the Jerk

Jerk is the derivative of acceleration with respect to time. Using \(a(t) = -\sin t - \cos t\), the jerk \(j(t)\) is \(\frac{da}{dt} = -\cos t + \sin t\). Substitute \(t=\frac{\pi}{4}\): \[ j\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0 \].

Key Concepts

Derivatives

In calculus, derivatives play a crucial role because they help us understand how functions change. A derivative of a function represents the rate of change or the slope of the function at a given point. For instance, if we have a position function, its derivative tells us the velocity.

• To find a derivative, you apply differentiation rules to the function in question.
• The derivative can give insights into how fast an object is moving or how quickly its speed is changing.

In our example, we have the position function \(s(t) = \sin t + \cos t\). By differentiating this, we find the velocity \(v(t)\). It's crucial to master this skill, as differentiation is used to find other important motion parameters such as acceleration and jerk.

Position Function

The position function \(s=f(t)\) describes the location of an object at any given time \(t\). It provides a map of where the object is positioned along a coordinate line. Knowing the exact position at different times can help us calculate how the object moves over time.

• Position is typically measured with respect to a reference point.
• Position functions can be constructed using different types of mathematical functions, such as polynomials, sine and cosine, etc.

In our scenario, the position is given by \(s(t) = \sin t + \cos t\), detailing how the object moves in meters over time in seconds.

Velocity

Velocity is a fundamental concept in understanding motion. It measures the rate of change of the position function with respect to time. In simpler terms, velocity tells us how fast the position of an object is changing and in which direction.

• Velocity is a vector, meaning it has both magnitude and direction.
• When velocity is positive, the object moves forward, and when negative, it moves backward.

To find the velocity of the object from our position function, we use the derivative: \(v(t) = \frac{ds}{dt} = \cos t - \sin t\). At the point \(t = \frac{\pi}{4}\), the velocity is 0, indicating the object is momentarily at rest.

Acceleration

Acceleration reflects how quickly an object's velocity changes over time. It's derived from the velocity function, making it the second derivative of the position function. Acceleration can inform us about the forces acting on an object.

• When acceleration is positive, the object's speed is increasing, while negative acceleration means the object is slowing down.
• Zero acceleration signifies constant velocity.

In this case, we find the acceleration by differentiating the velocity function: \(a(t) = \frac{dv}{dt} = -\sin t - \cos t\). At \(t = \frac{\pi}{4}\), the acceleration is \(-\sqrt{2}\), which indicates the object is slowing down at that moment.

Jerk

Jerk, also known as jolt, is the derivative of acceleration, informing us of the rate of change of acceleration. It's the third derivative of the position function and shows us how quickly acceleration is changing.

• Jerk is less commonly discussed but important in scenarios involving sudden starts or stops.
• High jerk values can lead to uncomfortable motion for passengers in vehicles, for example.

For our problem, jerk is calculated as \(j(t) = \frac{da}{dt} = -\cos t + \sin t\). At \(t = \frac{\pi}{4}\), the jerk is 0, suggesting no sudden change in acceleration at that particular moment.