Question

Sketch a position-time graph for an object that starts at \(x=1.5 \mathrm{~m}\), moves with a velocity of \(2.2 \mathrm{~m} / \mathrm{s}\) from \(t=0\) to \(t=1 \mathrm{~s}\), has a velocity of \(0 \mathrm{~m} / \mathrm{s}\) from \(t=1 \mathrm{~s}\) to \(t=2 \mathrm{~s}\), and has a velocity of \(-3.7 \mathrm{~m} / \mathrm{s}\) from \(t=2 \mathrm{~s}\) to \(t=5 \mathrm{~s}\).

Short answer

The graph is piecewise: increase, constant, then decrease.

Step-by-step solution

Initial Position and Set Up

Determine the initial position of the object, which is given as \( x = 1.5 \) m at time \( t = 0 \) s. Begin the graph at this point.

Calculate Position for First Interval

From \( t = 0 \) to \( t = 1 \) s, the object moves with a velocity of \( 2.2 \) m/s. The distance covered in this interval is \( d = v \cdot t = 2.2 \cdot 1 = 2.2 \) m. Add this to the initial position: \( x = 1.5 + 2.2 = 3.7 \) m. Plot the point \( (1, 3.7) \) on the graph.

Plot Constant Position for Second Interval

From \( t = 1 \) s to \( t = 2 \) s, the velocity is \( 0 \) m/s, meaning the position does not change. Plot a horizontal line from \( (1, 3.7) \) to \( (2, 3.7) \) on the graph.

Calculate Position for Third Interval

From \( t = 2 \) to \( t = 5 \) s, the object's velocity is \( -3.7 \) m/s. The distance covered is \( d = v \cdot t = -3.7 \cdot (5 - 2) = -11.1 \) m. Subtract this from the last known position: \( x = 3.7 - 11.1 = -7.4 \) m. Plot the point \( (5, -7.4) \) on the graph, drawing a line from \( (2, 3.7) \) to \( (5, -7.4) \).

Draw the Position-Time Graph

Use the points found and lines drawn in Steps 2-4 to sketch the position-time graph. The graph starts at \( (0, 1.5) \), rises to \( (1, 3.7) \), stays constant to \( (2, 3.7) \), and falls to \( (5, -7.4) \).

Key concepts

Understanding Velocity

Velocity is a measure of how quickly an object changes its position. It provides both the speed and direction of motion. In the exercise, the object's velocity changes across different time intervals.

Initially, with a velocity of \(2.2 \, \mathrm{m/s}\), the object moves forward, covering more distance over time. Later, with a velocity of \(0 \, \mathrm{m/s}\), the object remains stationary, indicating no change in position. Finally, the velocity shifts to \(-3.7 \, \mathrm{m/s}\), suggesting motion in the opposite direction.

This negative velocity denotes that the object moves backwards. Changes in velocity can be visually interpreted on a position-time graph by the slope of the graph line:

• Positive slopes indicate forward motion.
• Horizontal lines show no movement.
• Negative slopes denote motion in the opposite direction.
• The steeper the slope, the greater the speed.

Identifying Initial Position

The initial position of an object is where it starts on a position-time graph. It's key to determining where to begin tracking the object's movement.

In this scenario, the initial position is given as \(x = 1.5\) m at time \(t = 0\) seconds, meaning the object begins 1.5 meters along the measurement axis. This point acts as the starting reference for any distance covered in subsequent time intervals.

Every change in position is relative to this initial point. By knowing where an object starts, you can accurately plot its future positions by adding or subtracting the distance covered as time progresses.

Calculating Distance Covered

Distance covered refers to the actual path length traveled by the object over a time period.

It is calculated by multiplying velocity by time: \(d = v \cdot t\). Depending on the object's velocity (positive or negative), it will move forward or backward.

In the example provided:

• From \(t = 0\) to \(1 \, \mathrm{s}\), the distance is \(2.2 \, \mathrm{m}\).
• From \(t = 1\) to \(2 \, \mathrm{s}\), the distance is zero, meaning no movement.
• From \(t = 2\) to \(5 \, \mathrm{s}\), the distance covered is \(-11.1 \, \mathrm{m}\), which indicates reverse motion.

This calculated distance adjusts the position of the object on the graph at each time interval, revealing the trajectory.

Exploring Constant Position

A constant position on a position-time graph occurs when an object does not move during a period. This is represented by a horizontal line segment.

In real-world problems, an object might stay in the same place due to reasons like a halt or no applied force.For this exercise, from \(t = 1\) to \(2 \, \mathrm{s}\), the velocity is \(0 \, \mathrm{m/s}\), which keeps the object at a fixed position of \(3.7 \, \mathrm{m}\). This indicates that there's no distance traveled.

A flat, horizontal line on a graph signifies stillness, providing insight into periods of inactivity or balance in motion. Such segments on a graph are vital for understanding how and when the object's velocity changes over time.