Question

The equation of motion for a train on a straight track is \(x=11 \mathrm{~m}+(6.5 \mathrm{~m} / \mathrm{s})\) t. (a) Plot the position-time graph for the train from \(t=0\) to \(t=5.0 \mathrm{~s}\). (b) At what time is the train at \(x=32 \mathrm{~m}\) ?

Short answer

Time at 32 m is approximately 3.23 s.

Step-by-step solution

Understand the Equation of Motion

The given equation of motion for the train is a linear equation: \(x = 11\,\text{m} + (6.5\,\text{m/s})t\). The components of this equation include: 11 m as the initial position of the train, and 6.5 m/s as the constant velocity of the train.

Plot the Position-Time Graph

To plot the position-time graph, calculate the position of the train at different time intervals from \(t=0\) to \(t=5\) seconds using the equation \(x = 11 + 6.5t\). Generate data points: - When \(t=0, x=11\text{ m}\)- When \(t=1, x=17.5\text{ m}\)- When \(t=2, x=24\text{ m}\)- When \(t=3, x=30.5\text{ m}\)- When \(t=4, x=37\text{ m}\)- When \(t=5, x=43.5\text{ m}\).Then, plot these points on a graph with time on the x-axis and position on the y-axis, and draw a straight line through them.

Solve for Time When Train is at 32 m

To find the time when the position of the train is \(x = 32\,\text{m}\), use the equation \(11 + 6.5t = 32\). Solve for \(t\):\[ 6.5t = 32 - 11 \]\[ 6.5t = 21 \]\[ t = \frac{21}{6.5} \approx 3.23 \text{ s} \]

Key concepts

Position-Time Graph

A position-time graph visually represents the movement of an object over time. In this case, we are looking at a train's movement. The horizontal axis (x-axis) represents the time in seconds, while the vertical axis (y-axis) shows the position of the train in meters.
To construct this graph, begin by calculating the train's position at various time points using the equation of motion. For instance, when the time ( t ) is zero, the position ( x ) is 11 meters. Similarly, calculate for other time intervals, such as 1, 2, 3, 4, and 5 seconds, which give positions of 17.5 m, 24 m, 30.5 m, 37 m, and 43.5 m respectively.
Once you have these data points:

• (0, 11)
• (1, 17.5)
• (2, 24)
• (3, 30.5)
• (4, 37)
• (5, 43.5)

Plot them on the graph. Connecting these points will create a straight line, demonstrating linear motion of the train.

Linear Motion

Linear motion refers to movement along a straight path. This can be described by a linear equation, where the position changes at a constant rate over time. The term 'linear' signifies that a graph of this motion would be a straight line.
In the example provided, the train's motion is linear. The equation of motion, \( x = 11 + 6.5t \), clearly defines this linear path. Here, the real-world interpretation is that the train moves in a straight line along the track. Every second, the position increases by 6.5 meters, showing a uniform or unchanging motion pattern.
Linear motion is significant because it simplifies predictions. Once you understand the components of the equation of motion: the initial position (11 meters) and the constant velocity (6.5 m/s), you can easily predict where the train will be at any given time.

Constant Velocity

Constant velocity means that an object moves at a steady speed in a straight line. This concept is crucial in physics as it leads to predictable, uniform motion. Here, the train's constant velocity is 6.5 meters per second as indicated in the equation.
With constant velocity:

• Acceleration is zero because the speed doesn't increase or decrease.
• The position of the moving object changes uniformly over time.

In our train example, the speed doesn't change. Thus, for each second, it moves 6.5 meters further along the track. This kind of motion is idealized and assumes no external forces, such as friction or air resistance, affect the velocity.
Understanding constant velocity is essential when predicting future positions or plotting motion over time on a graph. It provides a fundamental expectation that guides more complex calculations.