Problem 5 - 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? $$ s=\frac{25}{t^{2}}-\frac{5}{t}, \quad 1 \leq t \leq 5 $$

Step-by-Step Solution

Answer

Displacement: -20 m, Average velocity: -5 m/s, No direction change during the interval.

Step 1: Calculate Displacement

To determine the displacement of the body, find the difference in position at the endpoints of the time interval. Compute the position at the start, \( s(1) \), and at the end, \( s(5) \), then take their difference:\[ s(1) = \frac{25}{1^2} - \frac{5}{1} = 25 - 5 = 20 \]\[ s(5) = \frac{25}{5^2} - \frac{5}{5} = \frac{25}{25} - 1 = 1 - 1 = 0 \]Displacement: \( s(5) - s(1) = 0 - 20 = -20 \text{ meters} \).

Step 2: Determine Average Velocity

Average velocity is the displacement divided by the time taken. Use the displacement from Step 1 and the total time interval (\( 5 - 1 = 4 \) seconds):\[ \text{Average Velocity} = \frac{-20}{4} = -5 \text{ m/s} \].

Step 3: Find Speed and Acceleration at Endpoints

Speed is the absolute value of velocity. First, find the derivative \( f'(t) \), which gives velocity:\[ f'(t) = -\frac{50}{t^3} + \frac{5}{t^2} \]Evaluate \( f'(t) \) at \( t = 1 \) and \( t = 5 \):\[ f'(1) = -50 + 5 = -45 \text{ m/s} \]\[ f'(5) = -\frac{50}{125} + \frac{5}{25} = -0.4 + 0.2 = -0.2 \text{ m/s} \]Thus, the speeds are \( |f'(1)| = 45 \text{ m/s} \) and \( |f'(5)| = 0.2 \text{ m/s} \).For acceleration, find the second derivative \( f''(t) \):\[ f''(t) = \frac{150}{t^4} - \frac{10}{t^3} \]Evaluate \( f''(t) \) at \( t = 1 \) and \( t = 5 \):\[ f''(1) = 150 - 10 = 140 \text{ m/s}^2 \]\[ f''(5) = \frac{150}{625} - \frac{10}{125} = 0.24 - 0.08 = 0.16 \text{ m/s}^2 \].

Step 4: Check for Change in Direction

A change in direction occurs when velocity changes sign. Solve for \( f'(t) = 0 \):\[ -\frac{50}{t^3} + \frac{5}{t^2} = 0 \]Factor out \( \frac{5}{t^3} \):\[ \frac{5}{t^3}(-10t + t) = 0 \rightarrow -10t + 1 = 0 \]Solve for \( t \):\[ 10t = 1 \rightarrow t = \frac{1}{10} \]Since \( \frac{1}{10} otin [1, 5] \), the body does not change direction during the interval.

Key Concepts

Displacement

Displacement is a vector quantity that represents the change in position of an object. It is calculated by finding the difference between the final and initial positions, along a straight line, within a specified time interval. In the exercise, we were given a function for the position of a body as a function of time, specifically \[ s(t) = \frac{25}{t^2} - \frac{5}{t}. \]To find the displacement, we compute the positions at the start and end of the time interval, which are when \( t = 1 \)and \( t = 5. \)This leads to:

• Initial position, \( s(1) = 20 \) meters
• Final position, \( s(5) = 0 \) meters

The displacement is calculated as:\[ s(5) - s(1) = 0 - 20 = -20 \, \text{meters}. \]This negative value indicates a change in position in the negative direction along the coordinate line. Remember, displacement is direction-aware, unlike distance, which is a scalar.

Average Velocity

Average velocity reports how quickly displacement occurs over a time interval, taking both magnitude and direction into account. It is calculated by dividing the total displacement by the total time taken. From the solution, we know the displacement is \(-20\) meters over a time interval from \(t = 1\) second to \(t = 5\) seconds.Thus, the total time is \(5 - 1 = 4\) seconds. Therefore, the average velocity is:\[\text{Average Velocity} = \frac{-20}{4} = -5 \, \text{meters per second (m/s)}. \]This negative average velocity indicates movement in the negative direction. It's important to recognize that average velocity gives us a general sense of the motion over the period rather than specific details about the speed at any given point.

Acceleration

Acceleration measures how the velocity of an object changes with time. It is a vector quantity expressed as the rate of change of velocity. In kinematics, it can be determined by computing the second derivative of the position function with respect to time. Here, the velocity \( f'(t) \) was first obtained as:\[ f'(t) = -\frac{50}{t^3} + \frac{5}{t^2}, \]and the acceleration \( f''(t) \) was computed as:\[ f''(t) = \frac{150}{t^4} - \frac{10}{t^3}. \]By evaluating this at the endpoints \( t = 1 \)and \( t = 5 \),

• Initial acceleration \( f''(1) = 140 \text{ m/s}^2 \)
• Final acceleration \( f''(5) = 0.16 \text{ m/s}^2\)

These values describe how quickly or slowly the body is changing its velocity at these two points. High acceleration at \( t = 1 \)suggests a rapidly changing velocity, while a smaller value at \( t = 5 \)indicates a more stable or reduced change in motion.

Velocity

Velocity is the speed of an object in a particular direction. Like displacement, it's a vector and is essential in understanding motion. It can be determined by taking the derivative of the position function with respect to time.The exercise involves calculating the velocity at specific points by using the derivative \( f'(t) \):

• At \( t = 1 \), the velocity is \( f'(1) = -45 \text{ m/s} \, \text{(meters per second)} \)
• At \( t = 5 \), the velocity is \( f'(5) = -0.2 \text{ m/s} \)

Although both values are negative, illustrating movement in the opposite direction to the positive coordinate line, they differ significantly in magnitude, indicating a large initial velocity that decreases over the time interval. Understanding velocity helps in assessing how quickly the position of an object is changing and in which direction, at any given moment.