Guided Practice: Position-Time Graphs in Physics with Calculus

Q1. When time increases, what happens to the distance?

Background

Topic: Position-Time Graphs

This question tests your understanding of how an object's position changes as time progresses, as represented on a position-time graph.

Key Terms:

• Position: The location of an object at a particular time, usually measured in meters (m).

• Time: The independent variable, usually measured in seconds (s).

• Distance: The total length of the path traveled by the object.

Step-by-Step Guidance

1. Look at the position-time graph and observe how the position (vertical axis) changes as you move to the right along the time (horizontal axis).

2. If the graph slopes upward, the position is increasing as time increases, indicating the object is moving away from the starting point.

3. If the graph is flat (horizontal), the position does not change as time increases, meaning the object is stationary.

4. Consider whether the graph ever slopes downward, which would indicate the object is moving back toward the starting point.

Try solving on your own before revealing the answer!

Final Answer:

As time increases, the distance (position) increases if the graph has a positive slope, stays the same if the graph is flat, or decreases if the graph has a negative slope.

This reflects the object's movement away from, remaining at, or moving toward the starting point.

Q2. Describe the object's motion.

Background

Topic: Interpreting Position-Time Graphs

This question asks you to interpret the shape and slope of a position-time graph to describe how the object is moving.

Key Terms:

• Slope: The rate at which position changes with time; represents velocity.

• Positive Slope: Object moves away from the origin.

• Negative Slope: Object moves toward the origin.

• Zero Slope: Object is stationary.

Step-by-Step Guidance

1. Examine the graph's slope: Is it positive, negative, or zero?

2. Note any changes in the slope: Does it become steeper or flatter at any point?

3. Describe the motion in terms of moving away from or toward the origin, or being stationary.

4. Summarize the object's motion over the entire time interval shown.

Try solving on your own before revealing the answer!

Final Answer:

The object's motion can be described by the slope of the graph: if the slope is positive, it moves away from the origin; if negative, it moves toward the origin; if zero, it is stationary. Any changes in slope indicate changes in speed or direction.

Q3. What is the velocity of the object?

Background

Topic: Calculating Velocity from Position-Time Graphs

This question tests your ability to determine the velocity of an object by analyzing the slope of a position-time graph.

Key Formula:

$v = \frac{\Delta x}{\Delta t}$

• $v$ = velocity (m/s)

• $\Delta x$ = change in position (meters)

• $\Delta t$ = change in time (seconds)

Step-by-Step Guidance

1. Identify two points on the position-time graph to determine the change in position ($\Delta x$) and change in time ($\Delta t$).

2. Calculate $\Delta x$ by subtracting the initial position from the final position.

3. Calculate $\Delta t$ by subtracting the initial time from the final time.

4. Set up the formula $v = \frac{\Delta x}{\Delta t}$ using your values for $\Delta x$ and $\Delta t$.

Try solving on your own before revealing the answer!

Final Answer:

The velocity is the slope of the position-time graph, calculated as $v = \frac{\Delta x}{\Delta t}$ using the chosen points.

Plug in the values from the graph to find the velocity in m/s.

Q4. When time increases, what happens to the distance?

Background

Topic: Position-Time Graphs

This is similar to Q1, focusing on interpreting how the object's position changes as time passes.

Key Terms:

• Position and Distance as described above.

Step-by-Step Guidance

1. Observe the direction of the graph's slope as time increases.

2. Determine if the position is increasing, decreasing, or remaining constant.

3. Relate this to the object's movement away from or toward the origin.

Try solving on your own before revealing the answer!

Final Answer:

As time increases, the distance (position) changes according to the slope: it increases with a positive slope, decreases with a negative slope, or stays the same with zero slope.

Q5. Describe the object's motion.

Background

Topic: Interpreting Position-Time Graphs

This question again asks you to describe the object's motion based on the graph's features.

Key Terms:

• See Q2 for definitions of slope and motion.

Step-by-Step Guidance

1. Analyze the graph for changes in slope or direction.

2. Describe whether the object is moving away, toward, or is stationary at different intervals.

3. Summarize the overall motion pattern.

Try solving on your own before revealing the answer!

Final Answer:

The object's motion is described by the slope: positive for moving away, negative for moving toward, and zero for stationary. Any changes in slope indicate changes in speed or direction.

Q6. What is the velocity of the object?

Background

Topic: Calculating Velocity from Position-Time Graphs

This is a repeat of Q3, focusing on determining velocity from the graph.

Key Formula:

$v = \frac{\Delta x}{\Delta t}$

• See Q3 for definitions.

Step-by-Step Guidance

1. Choose two points on the graph to find $\Delta x$ and $\Delta t$.

2. Calculate the change in position and time.

3. Set up the velocity formula and prepare to substitute the values.

Try solving on your own before revealing the answer!

Final Answer:

The velocity is found by dividing the change in position by the change in time: $v = \frac{\Delta x}{\Delta t}$.

Q7. Is the object's motion constant? Explain.

Background

Topic: Constant vs. Changing Velocity

This question tests your ability to determine whether the object moves at a constant velocity or if its speed/direction changes.

Key Terms:

• Constant Motion: The object moves at a steady speed in a straight line (slope does not change).

• Changing Motion: The object's speed or direction changes (slope changes).

Step-by-Step Guidance

1. Examine the graph for sections where the slope is constant (straight line) or changing (curved or with corners).

2. If the slope is the same throughout, the motion is constant.

3. If the slope changes, the motion is not constant.

4. Prepare to explain your reasoning based on the graph's features.

Try solving on your own before revealing the answer!

Final Answer:

The object's motion is constant if the graph's slope does not change; otherwise, it is not constant. The explanation should reference the graph's straight or changing slope.

Q8. What is the velocity during the first 4 seconds?

Background

Topic: Calculating Average Velocity from a Graph

This question asks you to find the average velocity over a specific time interval using the position-time graph.

Key Formula:

$v_{avg} = \frac{\Delta x}{\Delta t}$

• $\Delta x$ = change in position over the first 4 seconds

• $\Delta t$ = 4 seconds

Step-by-Step Guidance

1. Find the position at $t = 0$ and at $t = 4$ seconds on the graph.

2. Calculate $\Delta x$ by subtracting the initial position from the position at 4 seconds.

3. Set up the formula $v_{avg} = \frac{\Delta x}{4}$.

4. Prepare to substitute the values to find the average velocity.

Try solving on your own before revealing the answer!

Final Answer:

The average velocity during the first 4 seconds is $v_{avg} = \frac{\Delta x}{4}$, where $\Delta x$ is the change in position from $t=0$ to $t=4$ seconds.

Q9. What is the velocity during the last 6 seconds?

Background

Topic: Calculating Average Velocity Over a Time Interval

This question is similar to Q8 but focuses on the last 6 seconds of the motion.

Key Formula:

$v_{avg} = \frac{\Delta x}{\Delta t}$

• $\Delta x$ = change in position during the last 6 seconds

• $\Delta t$ = 6 seconds

Step-by-Step Guidance

1. Find the position at the start and end of the last 6 seconds (e.g., at $t = 4$ s and $t = 10$ s if the total time is 10 s).

2. Calculate $\Delta x$ for this interval.

3. Set up the formula $v_{avg} = \frac{\Delta x}{6}$.

4. Prepare to substitute the values to find the average velocity.

Try solving on your own before revealing the answer!

Final Answer:

The average velocity during the last 6 seconds is $v_{avg} = \frac{\Delta x}{6}$, where $\Delta x$ is the change in position during that interval.

Q10a. What is the velocity of the car for the first 3 seconds?

Background

Topic: Calculating Velocity from Displacement and Time

This question asks you to calculate the car's velocity during the first segment of its motion.

Key Formula:

$v = \frac{\Delta x}{\Delta t}$

• $\Delta x = 6$ meters (distance traveled in first 3 seconds)

• $\Delta t = 3$ seconds

Step-by-Step Guidance

1. Identify the displacement ($\Delta x$) and time interval ($\Delta t$) for the first 3 seconds.

2. Set up the formula $v = \frac{6}{3}$.

3. Prepare to perform the division to find the velocity.

Try solving on your own before revealing the answer!

Final Answer:

The velocity for the first 3 seconds is $v = \frac{6}{3} = 2$ m/s.

Q10b. What is the velocity of the car from times 3-8 seconds?

Background

Topic: Velocity During a Stationary Interval

This question asks you to find the velocity while the car is stopped.

Key Formula:

$v = \frac{\Delta x}{\Delta t}$

• $\Delta x = 0$ (no movement)

• $\Delta t = 5$ seconds (from 3 to 8 seconds)

Step-by-Step Guidance

1. Recognize that the car is stopped, so its position does not change.

2. Set up the formula $v = \frac{0}{5}$.

3. Prepare to perform the division to find the velocity.

Try solving on your own before revealing the answer!

Final Answer:

The velocity from 3 to 8 seconds is $v = 0$ m/s, since the car is not moving during this interval.

Q10c. During which time is the car moving faster, 0-3s or 8-10s? How could you know this without calculating the velocity?

Background

Topic: Comparing Speeds Using Graph Slope

This question asks you to compare the car's speed during two intervals by analyzing the graph's slope.

Key Terms:

• Slope: Steeper slope means higher speed.

Step-by-Step Guidance

1. Look at the graph segments for 0-3s and 8-10s.

2. Compare the steepness of the slopes in these intervals.

3. The steeper the slope, the faster the car is moving.

4. Prepare to state which interval has the steeper slope.

Try solving on your own before revealing the answer!

Final Answer:

The car is moving faster during the interval with the steeper slope. You can tell this without calculation by visually comparing the slopes on the graph.

Q11a. What is the velocity of the car for the first 2 seconds?

Background

Topic: Calculating Velocity from Displacement and Time

This question asks you to calculate the car's velocity during the first 2 seconds of motion.

Key Formula:

$v = \frac{\Delta x}{\Delta t}$

• $\Delta x = 8$ meters

• $\Delta t = 2$ seconds

Step-by-Step Guidance

1. Identify the displacement and time interval for the first 2 seconds.

2. Set up the formula $v = \frac{8}{2}$.

3. Prepare to perform the division to find the velocity.

Try solving on your own before revealing the answer!

Final Answer:

The velocity for the first 2 seconds is $v = \frac{8}{2} = 4$ m/s.

Q11b. What is the velocity of the car from times 2-5 seconds?

Background

Topic: Velocity During a Stationary Interval

This question asks you to find the velocity while the car is not moving.

Key Formula:

$v = \frac{\Delta x}{\Delta t}$

• $\Delta x = 0$ (no movement)

• $\Delta t = 3$ seconds (from 2 to 5 seconds)

Step-by-Step Guidance

1. Recognize that the car is stationary during this interval.

2. Set up the formula $v = \frac{0}{3}$.

3. Prepare to perform the division to find the velocity.

Try solving on your own before revealing the answer!

Final Answer:

The velocity from 2 to 5 seconds is $v = 0$ m/s, since the car is not moving during this interval.

Q11c. What is the velocity of the car from times 5-10s?

Background

Topic: Calculating Velocity with Negative Displacement

This question asks you to calculate the velocity when the car moves in the negative direction.

Key Formula:

$v = \frac{\Delta x}{\Delta t}$

• $\Delta x = -5$ meters (negative direction)

• $\Delta t = 5$ seconds (from 5 to 10 seconds)

Step-by-Step Guidance

1. Identify the displacement and time interval for 5-10 seconds.

2. Set up the formula $v = \frac{-5}{5}$.

3. Prepare to perform the division to find the velocity (note the negative sign).

Try solving on your own before revealing the answer!

Final Answer:

The velocity from 5 to 10 seconds is $v = \frac{-5}{5} = -1$ m/s. The negative sign indicates motion in the opposite direction.

Q12a. What is the velocity of the car for the first 2 seconds?

Background

Topic: Calculating Velocity from Displacement and Time

This question asks you to calculate the car's velocity during the first 2 seconds of motion.

Key Formula:

$v = \frac{\Delta x}{\Delta t}$

• $\Delta x = 5$ meters

• $\Delta t = 2$ seconds

Step-by-Step Guidance

1. Identify the displacement and time interval for the first 2 seconds.

2. Set up the formula $v = \frac{5}{2}$.

3. Prepare to perform the division to find the velocity.

Try solving on your own before revealing the answer!

Final Answer:

The velocity for the first 2 seconds is $v = \frac{5}{2} = 2.5$ m/s.

Q12b. What is the velocity of the car from times 4-8 seconds?

Background

Topic: Calculating Velocity with Negative Displacement

This question asks you to calculate the velocity when the car moves in the opposite direction.

Key Formula:

$v = \frac{\Delta x}{\Delta t}$

• $\Delta x = -5$ meters (opposite direction)

• $\Delta t = 4$ seconds (from 4 to 8 seconds)

Step-by-Step Guidance

1. Identify the displacement and time interval for 4-8 seconds.

2. Set up the formula $v = \frac{-5}{4}$.

3. Prepare to perform the division to find the velocity (note the negative sign).

Try solving on your own before revealing the answer!

Final Answer:

The velocity from 4 to 8 seconds is $v = \frac{-5}{4} = -1.25$ m/s. The negative sign indicates motion in the opposite direction.

Q12c. How far did the car move from 8-10s?

Background

Topic: Calculating Distance from Velocity and Time

This question asks you to find the distance traveled during a specific interval, given the velocity and time.

Key Formula:

$\Delta x = v \cdot \Delta t$

• $v = 3$ m/s (given in the problem)

• $\Delta t = 2$ seconds (from 8 to 10 seconds)

Step-by-Step Guidance

1. Identify the velocity and time interval for 8-10 seconds.

2. Set up the formula $\Delta x = 3 \times 2$.

3. Prepare to perform the multiplication to find the distance.

Try solving on your own before revealing the answer!

Final Answer:

The car moved $\Delta x = 3 \times 2 = 6$ meters from 8 to 10 seconds.