Calculating Displacement from Velocity-Time Graphs: Videos & Practice Problems

In a velocity-time graph, displacement (ΔX) is determined by calculating the area under the curve between two time points. This area can be broken down into simpler shapes like rectangles and triangles, using the formulas: rectangle area = base × height and triangle area = ½ × base × height. For example, if the area above the time axis is positive and below is negative, total displacement is the sum of these areas. Understanding this concept is crucial for analyzing motion and applying the principles of kinematics effectively.

Calculating Displacement from Velocity-Time Graphs

In physics, calculating displacement using a velocity-time graph is a fundamental skill. Displacement, represented as ΔX, is determined by finding the area under the curve of the graph between two time points. This area corresponds to the change in position of an object over time. The term "under the curve" refers to the space between the velocity values and the time axis, which can be visualized as geometric shapes such as rectangles and triangles.

To calculate displacement, one can break down complex shapes into simpler ones. For instance, the area of a rectangle is calculated using the formula:

Area of Rectangle = Base × Height

For triangles, the area is given by:

Area of Triangle = \(\frac{1}{2} \times \text{Base} \times \text{Height}\)

Consider a velocity-time graph where you need to find the displacement over a specific interval, such as the first 4 seconds. By identifying the relevant area, you can calculate the displacement as follows: if the area consists of a rectangle with a base of 2 meters and a height of 2 meters, the area would be:

Area = 2 m × 2 m = 4 m²

For a triangle with the same dimensions, the area would be:

Area = \(\frac{1}{2} \times 2 m \times 2 m = 2 m²\)

Adding these areas together gives a total displacement of:

ΔX = 4 m + 2 m = 6 m

When calculating total displacement over a longer interval, such as from 0 to 6 seconds, it is essential to consider both positive and negative areas. Areas above the time axis indicate positive displacement, while areas below indicate negative displacement. For example, if the area from 4 to 6 seconds is a triangle with a base of 2 meters and a height of 2 meters, the area would initially be calculated as:

Area = \(\frac{1}{2} \times 2 m \times 2 m = 2 m²\)

However, since this area is below the time axis, it is treated as negative, resulting in:

ΔX4 = -2 m

Thus, the total displacement from 0 to 6 seconds would be:

Total ΔX = 6 m + (-2 m) = 4 m

Understanding how to interpret and calculate areas under a velocity-time graph is crucial for analyzing motion, as it allows for the determination of displacement in various scenarios.

A moving box's motion is described by the velocity graph below. The box's initial position is x = 0.
(a) How far has the box moved at t = 4?
(b) What is the displacement of the box from t=4 to t=10?

• A. (a) 16 m;
  (b) 0 m
• B. (a) 16 m;
  (b) 24 m
• C. (a) 24 m;
  (b) 0 m
• D. (a) 24 m;
  (b) 24 m
• E. (a) 32 m;
  (b) 0 m
• F. (a) 32 m;
  (b) 24 m

Answer: (C)

Displacement of a Car

To determine the final position of a moving car using a velocity-time graph, we start with the initial position and calculate the displacement. The initial position is given as -21 meters. The displacement, denoted as Δx, can be calculated using the formula:

Δx = x_final - x_initial

In the context of a velocity-time graph, the displacement is represented by the area under the curve between the specified time intervals. Therefore, we can rearrange the formula to find the final position:

x_final = x_initial + Δx

To find Δx, we need to calculate the area under the curve from t = 0 seconds to t = 5 seconds. This area can be broken down into simpler geometric shapes. In this case, we identify three shapes: a large triangle, a rectangle, and a smaller triangle.

1. **Area of the first triangle (Δx₁):** The base is 3 seconds (from t = 0 to t = 3) and the height is 8 m/s (from 2 m/s to 10 m/s). The area is calculated as:

Δx₁ = ½ × base × height = ½ × 3 × 8 = 12 m

2. **Area of the rectangle (Δx₂):** The base is 5 seconds (from t = 0 to t = 5) and the height is 2 m/s. The area is:

Δx₂ = base × height = 5 × 2 = 10 m

3. **Area of the smaller triangle (Δx₃):** The base is 2 seconds (from t = 3 to t = 5) and the height is 2 m/s (from 2 m/s to 4 m/s). The area is:

Δx₃ = ½ × base × height = ½ × 2 × 2 = 2 m

Now, we sum the areas to find the total displacement:

Δx = Δx₁ + Δx₂ + Δx₃ = 12 + 10 + 2 = 24 m

Finally, we substitute the displacement back into the equation for final position:

x_final = x_initial + Δx = -21 + 24 = 3 m

Thus, the final position of the car at t = 5 seconds is 3 meters.

Here’s what students ask on this topic:

How do you calculate displacement from a velocity-time graph?

To calculate displacement from a velocity-time graph, you need to find the area under the curve between two time points. This area represents the displacement (ΔX). You can break the area into simpler shapes like rectangles and triangles. For a rectangle, use the formula: $\mathrm{area}=\mathrm{base}\times \mathrm{height}$. For a triangle, use: $\mathrm{area}=\frac{1}{2}\mathrm{base}\times \mathrm{height}$. Sum these areas to get the total displacement. Remember, areas above the time axis are positive, and areas below are negative.

What does the area under a velocity-time graph represent?

The area under a velocity-time graph represents the displacement (ΔX) of an object over a given time interval. This is because displacement is the integral of velocity with respect to time. By calculating the area under the curve, you are effectively summing up the product of velocity and time, which gives you the total change in position. Positive areas indicate forward motion, while negative areas indicate backward motion.

How do you find the displacement if the velocity-time graph has both positive and negative areas?

To find the displacement when a velocity-time graph has both positive and negative areas, calculate the area of each section separately. Positive areas (above the time axis) represent forward displacement, while negative areas (below the time axis) represent backward displacement. Sum these areas, taking care to add positive areas and subtract negative areas. The result is the net displacement. For example, if you have +6 m and -2 m, the total displacement is 4 m.

What is the formula for the area of a triangle on a velocity-time graph?

The formula for the area of a triangle on a velocity-time graph is: $\mathrm{area}=\frac{1}{2}\mathrm{base}\times \mathrm{height}$. Here, the base is the time interval, and the height is the velocity. This formula helps you calculate the area under the curve when the section of the graph forms a triangular shape, contributing to the total displacement.

Why is it important to understand displacement from velocity-time graphs?

Understanding displacement from velocity-time graphs is crucial because it allows you to analyze and interpret the motion of objects accurately. This skill is essential in kinematics, helping you solve problems related to motion in physics. By mastering this concept, you can determine how far an object has traveled over a period, which is fundamental for understanding real-world applications like vehicle movement, projectile motion, and more.