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Velocity Time Graphs – Definition, Examples

Aug 20, 2022
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Key Concepts

1. Velocity-time graph

2. Slope of v-t graph = acceleration

3. Types of velocity-time graphs

4. Displacement from v-t graphs

Introduction

The velocity of a body in a uniformly accelerated motion increases by equal amounts in equal intervals of time. This also indicates that it moves at a constant acceleration. When its velocity at different instants of time is plotted against the corresponding values of time, the graph turns out to be a straight line that passes through the origin (if the body’s motion is considered to start from rest) and is inclined to the x-axis (time-axis). Such a graph, when plotted for a moving body, can provide a lot of information about the motion of the body, such as the type of motion, velocity, acceleration, and displacement. 

parallel

Explanation:

Acceleration from a v-t graph: 

The slope is a measure of the steepness of a graph. Consider two points A and B in the position-time graph of a body in a uniform motion and let their coordinates be (t1, v1) and (t2, v2), respectively, as shown in the figure below. 

 Slope of a v-t graph 

The slope of this graph is given by, 

Slope, m = BC/ AC 

Or, m = (v2 – v1)/ (t2 – t1) 

parallel

In fact, the slope of a v-t graph at a particular instant gives the acceleration of the body at that instant.  

This is because the slope is the ratio of the change in velocity and change in time, which is equal to the velocity of the body. 

 Or, v = m = (v2 – v1)/ (t2 – t1) 

Types of velocity-time graphs: 

The velocity-time graph may look very different for different kinds of motion. Some of them are as follows: 

  1. Velocity-time graph of a body in a uniform motion: 

An object moving at a uniform velocity covers equal distances in equal intervals of time. This is because its velocity remains constant throughout the motion. Such a motion is called a uniform motion. Its v-t graph is a straight line parallel to the x-axis, as shown in the graph below. The slope of the velocity-time graph of a body in uniform motion is zero. This indicates that the acceleration of the body in such a motion is zero

 v-t graph for uniform motion] 

  1. Velocity-time graph of a uniformly and positively accelerating body: 

An object in a uniformly and positively accelerated motion increases its velocity by equal amounts in equal intervals of time. This is because its acceleration remains constant throughout its motion. The v-t graph of such a motion, when plotted, turns out to be a straight line passing through the origin (if the body starts from rest) with a positive slope. The positive and constant slope indicates the positive and constant acceleration of the body. 

v-t graph of a uniformly and positively accelerated motion

  1. Velocity-time graph of a uniformly and negatively accelerating body: 

An object in a uniformly and negatively accelerated motion decreases its velocity by equal amounts in equal intervals of time. This is because its acceleration remains constant but negative throughout its motion. The v-t graph of such a motion, when plotted, turns out to be a straight line with a negative slope. The negative and constant slope indicates the negative and constant acceleration of the body. 

v-t graph of a uniformly and negatively accelerated motion 

Comparing accelerations: 

The accelerations of different bodies can be compared by looking at the steepness (slope) of their v-t graphs. For example, the graph below shows the v-t graph of two bodies in uniform motion at different accelerations. It is easy to figure out from the graph that the acceleration of the first body is greater than the second. 

Comparing accelerations

  1. Velocity-time graph of a body at rest: 

The velocity of a body at rest is zero until it starts moving. Therefore, the v-t graph of a body at rest is a straight line coinciding with the x-axis (time-axis), as shown in the figure below. 

v-t graph of a body at rest 

  

  1. Velocity-time graph of a body with increasing acceleration: 

In this type of motion, the magnitude of the acceleration of a body keeps increasing with time. The plot of velocity and time for such a motion is curved in a way shown in the figure below. The graph would start from the origin if the body was initially at rest. Else, it may start from a point on the x or y-axis or any other point on the cartesian plane. The slope of the v-t graph of a body moving with increasing acceleration keeps increasing as the steepness of the graph keeps increasing. 

 v-t graph of a body with an increasing acceleration

  1. Velocity-time graph of a body with decreasing acceleration: 

In this type of motion, the magnitude of the acceleration of a body keeps decreasing with time. The plot of velocity and time for such a motion is curved in a way shown in the figure below. The graph would start from the origin if the body was initially at rest. Else, it may start from a point on the x or y-axis or any other point on the cartesian plane. The slope of the v-t graph of a body moving with decreasing acceleration keeps decreasing as the steepness of the graph keeps decreasing. 

 Variation of slope 

Displacement from a velocity-time graph: 

The displacement covered by a body is equal to the area under its velocity-time graph. In the velocity-time graph given below, let the coordinates of A and B be (t, v) and (t, 0), respectively. 

Total displacement covered by the body = the area under the v-t graph (OA)  

                                                                              = area of the triangle OAB. 

Therefore,  

Total displacement = area(triangle OAB) 

                                 = (1/2) x AB x OB 

                                 = (1/2)vt 

Displacement = area under the graph 

The v-t graph might also look different in other cases. Or we might need to find the displacement of the body in a section of the graph. Even in those cases, the area under the graph gives the displacement of the body. Here is another example. 

Displacement = area under the v-t graph  = area (trapezium OABD) 

Or, s = area (ABC) + area (OACD) 

Or, s = (1/2) x AC x BC + AC x CD 

Or, s = (1/2) t2(v2 – v1) + t2v1 

Or, s = t2 [(1/2) (v2 – v1) + v1] 

If the displacement of a portion of the motion is required, then the area under the corresponding portion of the v-t graph should be calculated. 

Displacement = area under the graph 

Problems 

  1. Identify the type of acceleration (positive, negative, zero, and non-uniform) the body is undergoing by analyzing every section of the velocity-time graph shown below.​ 
Question 

Answer: 

  1. Non-uniformly accelerated motion  
  1. Uniformly accelerated motion  
  1. Uniform motion  
  1. Uniformly decelerated motion  
  1. Uniformly accelerated motion  
  1. Non-uniformly decelerated motion  
  1. Identify the type of motion (uniform, non-uniform, uniformly, and non-uniformly accelerated/ decelerated motion) the body is undergoing by analyzing every section of the velocity-time graph shown below.​ 

Answer: 

  1. Non-uniformly accelerated motion  
  1. Uniformly accelerated motion  
  1. Uniform motion  
  1. Uniformly decelerated motion  
  1. Uniformly accelerated motion  
  1. Non-uniformly decelerated motion  
  1. The v-t graph of a body’s motion is shown in the figure below. Calculate its total displacement. 

Solution: 

Total displacement = area (OABD) + area (BDC) 

                                 = AB BD + (1/2) BD DC 

                                 = 3 x 6 + (1/2) x 6 x (7 – 3) 

                                 = 18 + (1/2) x 6 x 4 

                                 = 18 + 12 

                                 = 30 m 

  1. The v-t graph of a body’s motion is shown in the figure below. Calculate the acceleration of the body in each section of the graph. In which section does the body have the greatest acceleration? 

Solution: 

SECTION – A 

Consider the end-points on the v-t graph with coordinates (0,0) and (25, 50). 

Acceleration = slope = (50 – 0)/ (25 – 0) 

                                    = 50/ 25 

                                    = 2 m/s2 

SECTION – B 

Consider the end-points on the v-t graph with coordinates (25, 50) and (60, 50). 

Acceleration = slope = (50 – 50)/ (60 – 25) 

                                    = 0 

SECTION – C 

Consider the end-points on the v-t graph with coordinates (60, 50) and (70, 80). 

Acceleration = slope = (80 – 50)/ (70 – 60) 

                                    = 30/ 10 

                                    = 3 m/s2 

The body has the greatest acceleration in section C. 

Summary

1. In the velocity-time graph the velocity (y-axis) of a body is plotted against the time (x-axis) of a body.

2. The v-t graph for a motion with uniformly increasing velocity is a straight line with a positive slope.

3. The v-t graph for a motion with a uniformly decreasing velocity is a straight line with a negative slope.

4. The acceleration of a moving body is equal to the slope of its v-t graph.

5. Greater the slope of a v-t graph, greater the acceleration of The body.

6. The v-t graph of a body at rest coincides with die x-axis.

7. The v-t graph of a body with an increasing/decreasing acceleration is a curve whose slope keeps increasing/decreasing as the time increases.

8. The displacement of a body is equal to the area under the v-t graph of a body.

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