Guidance for 1-D Kinematics Graphical Methods: Calculating Displacement from Velocity-Time Graphs

Q7.5: The graph represents the velocity-time plot of a straight-line motion of a car. How far does the car travel between $t = 2\ \text{s}$ and $t = 9\ \text{s}$?

Background

Topic: 1-D Kinematics – Graphical Methods

This question tests your ability to interpret a velocity-time graph and calculate the displacement (distance traveled) over a specific time interval. The area under the velocity-time curve between two points gives the displacement.

Key Terms and Formulas

• Velocity ($v$): The rate of change of position with respect to time.

• Displacement ($\Delta x$): The change in position, found by integrating (or summing) the area under the velocity-time graph.

• Area under the curve: For straight-line segments, calculate the area of rectangles and triangles.

Key formula:

$\Delta x = \int_{t_1}^{t_2} v(t)\,dt$

For piecewise linear graphs, sum the areas of geometric shapes (rectangles, triangles) between $t_1$ and $t_2$.

Step-by-Step Guidance

1. Examine the velocity-time graph between $t = 2\ \text{s}$ and $t = 9\ \text{s}$. Identify the shapes formed under the curve in this interval (rectangles and triangles).

2. Calculate the area under the curve for each segment. For example, from $t = 2\ \text{s}$ to $t = 5\ \text{s}$, the velocity is constant, forming a rectangle. From $t = 5\ \text{s}$ to $t = 9\ \text{s}$, the velocity decreases linearly, forming a triangle.

3. For the rectangle: Area = width × height. For the triangle: Area = $\frac{1}{2}$ × base × height.

4. Add the areas together to find the total displacement between $t = 2\ \text{s}$ and $t = 9\ \text{s}$.

Try solving on your own before revealing the answer!

Final Answer: 48 m

$\Delta x = (5 - 2) \times 12 + \frac{1}{2} \times (9 - 5) \times 12 = 36 + 24 = 48\ \text{m}$

The total displacement is the sum of the rectangle and triangle areas under the velocity-time curve.