How to Find Velocity from Displacement: A Comprehensive Guide

Velocity is a fundamental concept in physics that describes the rate of change of an object’s position over time. To find the velocity of an object from its displacement, you can use various techniques, including the formula, calculus, and graphical analysis. This comprehensive guide will walk you through the step-by-step process of determining velocity from displacement, providing you with a deep understanding of the underlying principles and practical applications.

Understanding the Velocity-Displacement Relationship

The relationship between velocity and displacement is governed by the following formula:

$$v = \frac{s}{t}$$

where:
– v is the velocity of the object
– s is the displacement of the object
– t is the time taken to travel the displacement

This formula indicates that velocity is the ratio of the displacement to the time taken to cover that displacement. The displacement is the change in the object’s position, and the time is the duration over which this change occurs.

It’s important to note that displacement is a vector quantity, meaning it has both magnitude and direction. This means that the velocity calculated using the formula above will also be a vector quantity, with both magnitude and direction.

Calculating Velocity from Displacement: Algebraic Approach

1. Constant Velocity: If the object is moving at a constant velocity, you can use the formula directly to calculate the velocity. For example, if an object travels a distance of 100 meters in 10 seconds, the velocity would be:

$$v = \frac{s}{t} = \frac{100 \, \text{m}}{10 \, \text{s}} = 10 \, \text{m/s}$$

1. Displacement as a Function of Time: If the displacement of the object is given as a function of time, s(t), you can find the velocity by taking the derivative of the displacement function with respect to time. This is because velocity is the rate of change of displacement with respect to time.

For instance, if the displacement function is s(t) = 5t^2, the velocity function would be:

$$v(t) = \frac{ds}{dt} = \frac{d}{dt}(5t^2) = 10t$$

In this case, the velocity is a function of time, meaning it changes as the object moves.

Graphical Approach to Finding Velocity from Displacement

If the displacement of the object is given as a graph, you can find the velocity by analyzing the slope of the graph.

1. Straight-Line Displacement-Time Graph: If the displacement-time graph is a straight line, the slope of the line represents the constant velocity of the object. For example, if the displacement-time graph has a slope of 5 m/s, the velocity of the object is 5 m/s.

2. Curved Displacement-Time Graph: If the displacement-time graph is a curve, the velocity is the slope of the tangent line to the curve at a specific point. The tangent line is the line that touches the curve at that point and has the same slope as the curve at that point.

For a parabolic displacement-time graph, the velocity is the slope of the tangent line at a particular point on the curve.

Practical Examples and Numerical Problems

1. Example 1: An object travels a distance of 50 meters in 5 seconds. Calculate the velocity of the object.

Solution:
$$v = \frac{s}{t} = \frac{50 \, \text{m}}{5 \, \text{s}} = 10 \, \text{m/s}$$

1. Example 2: The displacement of an object is given by the function s(t) = 3t^2 + 2t, where t is in seconds and s is in meters. Find the velocity of the object at t = 3 seconds.

Solution:
The velocity is the derivative of the displacement function with respect to time:
$$v(t) = \frac{ds}{dt} = \frac{d}{dt}(3t^2 + 2t) = 6t + 2$$
Substituting t = 3 seconds, we get:
$$v(3) = 6(3) + 2 = 20 \, \text{m/s}$$

1. Example 3: The displacement-time graph of an object is a parabola. Sketch the velocity-time graph for this object.

Solution:
Since the displacement-time graph is a parabola, the velocity-time graph will be a straight line. The slope of the velocity-time graph will be the derivative of the displacement function, which is a linear function.

1. Numerical Problem 1: An object moves along a straight line with a displacement function s(t) = 2t^3 - 3t^2 + 4t, where s is in meters and t is in seconds. Calculate the velocity of the object at t = 2 seconds.

Solution:
The velocity is the derivative of the displacement function with respect to time:
$$v(t) = \frac{ds}{dt} = \frac{d}{dt}(2t^3 – 3t^2 + 4t) = 6t^2 – 6t + 4$$
Substituting t = 2 seconds, we get:
$$v(2) = 6(2)^2 – 6(2) + 4 = 24 – 12 + 4 = 16 \, \text{m/s}$$

1. Numerical Problem 2: The displacement-time graph of an object is a straight line with a slope of 8 m/s. Find the velocity of the object.

Solution:
Since the displacement-time graph is a straight line, the slope of the line represents the constant velocity of the object. Therefore, the velocity of the object is 8 m/s.

These examples and numerical problems demonstrate the various techniques and applications of finding velocity from displacement, covering both algebraic and graphical approaches. By working through these examples, you can develop a deeper understanding of the concepts and improve your problem-solving skills in this area of physics.

Additional Resources and References

1. Understanding Displacement and Velocity
2. How to Calculate Velocity from a Displacement Time Graph
3. Instantaneous and Average Velocity – HSC Physics
4. How To Find The Velocity From a Displacement-Time Graph

Remember, understanding the relationship between displacement and velocity is crucial in various fields of physics, from classical mechanics to modern physics. By mastering the techniques presented in this guide, you’ll be well-equipped to tackle a wide range of problems and deepen your understanding of this fundamental concept.