Problem 2 - Thomas Calculus

Question

Exercises \(1-6\) give the positions \(s=f(t)\) of a body moving on a coordinate line, with \(s\) in meters and \(t\) in seconds. a. Find the body's displacement and average velocity for the given time interval. b. Find the body's speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? $$ s=6 t-t^{2}, \quad 0 \leq t \leq 6 $$

Step-by-Step Solution

Answer

Displacement is 0 meters, average velocity is 0 m/s. Speed is 6 m/s at endpoints, acceleration is -2 m/s². Direction changes at t = 3 seconds.

Step 1: Understand Displacement

Displacement is the change in position of the body over the given time interval. It's found by evaluating the position function at the endpoints and taking the difference.\[\text{Displacement} = f(t_2) - f(t_1)\]For the interval \([0, 6]\), evaluate \(s = 6t - t^2\) at \(t=0\) and \(t=6\).

Step 2: Calculate Displacement

Evaluate the function at the endpoints:\[f(0) = 6(0) - 0^2 = 0\]\[f(6) = 6(6) - 6^2 = 36 - 36 = 0\]Now, calculate the displacement:\[\text{Displacement} = 0 - 0 = 0\, \text{meters}\]

Step 3: Understand Average Velocity

Average velocity is the displacement divided by the time interval:\[\text{Average velocity} = \frac{f(t_2) - f(t_1)}{t_2 - t_1}\]Here, since the displacement is 0 meters over 6 seconds, the average velocity follows.

Step 4: Calculate Average Velocity

Using the displacement (0 meters) over the time period (6 seconds):\[\text{Average velocity} = \frac{0}{6 - 0} = 0\, \text{m/s}\]

Step 5: Understand Speed and Acceleration

Speed is the magnitude of velocity, and acceleration is the rate of change of velocity.Velocity is the derivative of the position function:\[v(t) = \frac{ds}{dt} = \frac{d}{dt}(6t - t^2) = 6 - 2t\]Acceleration is the derivative of velocity:\[a(t) = \frac{dv}{dt} = \frac{d}{dt}(6 - 2t) = -2\]

Step 6: Calculate Speed at Endpoints

Evaluate the velocity at \(t=0\) and \(t=6\):\[v(0) = 6 - 2(0) = 6\, \text{m/s}\]\[v(6) = 6 - 2(6) = 6 - 12 = -6\, \text{m/s}\]Speed at \(t=0\) is \(|6| = 6\, \text{m/s}\) and at \(t=6\) is \(|-6| = 6\, \text{m/s}\).

Step 7: Calculate Acceleration at Endpoints

Since acceleration is constant:\[a(t) = -2\, \text{m/s}^2\]Acceleration remains \(-2\, \text{m/s}^2\) at both \(t=0\) and \(t=6\).

Step 8: Determine Change in Direction

A change in direction occurs when velocity changes sign. Set velocity function to zero:\[v(t) = 6 - 2t = 0\]\[2t = 6 \Rightarrow t = 3\]Check velocity signs around \(t=3\). At \(t=2\), \(v(2)=2\) (positive), and at \(t=4\), \(v(4)=-2\) (negative). Thus, the body changes direction at \(t=3\, \text{seconds}\).

Key Concepts

Understanding Displacement

In kinematics, displacement is a vital concept. It measures how far an object has moved from its original position over a specific time interval. Unlike distance, displacement is a vector quantity, meaning it takes into account not just how much an object has moved, but in what direction. For the exercise at hand, displacement is calculated by evaluating the position function at the two endpoints of the time interval and finding the difference between them.

• The position function given is \(s = 6t - t^2\) for the interval \([0, 6]\).
• Evaluating at \(t = 0\), we get: \(f(0) = 6(0) - 0^2 = 0\).
• At \(t = 6\), we find: \(f(6) = 6(6) - 6^2 = 36 - 36 = 0\).

Thus, the displacement is \(0 - 0 = 0\) meters, indicating there is no net change in position.

Grasping Velocity

Velocity defines the speed of an object in a particular direction. It's another vector quantity, crucial in understanding how fast an object moves and where it's heading. The average velocity is computed as the total displacement divided by the total time. In this example, we have:

• The equation for average velocity: \(\text{Average velocity} = \frac{\text{Displacement}}{t_2 - t_1}\).
• With a displacement of 0 meters over a period of 6 seconds, we calculate: \(\frac{0}{6 - 0} = 0\, \text{m/s}\).

This result means that, on average, the body's speed in the given time is zero, as it ends up at the starting point.

Exploring Acceleration

Acceleration tells us about how quickly the velocity of an object is changing. In kinematics, it's extremely helpful to determine if an object is speeding up, slowing down, or maintaining a constant speed. Acceleration can be found by differentiating the velocity function:

• Velocity, \(v(t)\), is the derivative of the position function: \(v(t) = \frac{d}{dt}(6t - t^2) = 6 - 2t\).
• Acceleration, \(a(t)\), is the derivative of velocity: \(a(t) = \frac{d}{dt}(6 - 2t) = -2\).

Hence, the acceleration is \(-2\, \text{m/s}^2\) constantly throughout the interval, signifying a consistent decrease in velocity.

Identifying Direction Change

Direction changes are crucial in kinematics, indicating where an object reverses its course. A change in direction occurs when the velocity changes its sign — from positive to negative or vice versa. To find when a body changes direction:

• Set the velocity equation to zero: \(v(t) = 6 - 2t = 0\).
• Solving gives \(2t = 6 \Rightarrow t = 3\).
• Check velocity just before and after \(t = 3\):
• At \(t = 2\), velocity is positive (\(v(2) = 2\)).
• At \(t = 4\), velocity is negative (\(v(4) = -2\)).

This confirms that the body indeed changes direction at \(t = 3\) seconds, shifting from moving positively along the path to negatively.