Graphs: Introduction, Comparison

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Graphs are a way of expressing the relationship between two values, a change in one value results in a change in the other counterpart. It is a representation of data in a two-dimensional diagram that describes the relationship between dependent and independent variables. Independent variables are represented in a horizontal line known as the X-axis while the dependent variants are represented in a vertical line known as the Y-axis. In Physics, it is effective to interpret motions by observing graphs. Here, we will be discussing the topic in detail along with some important questions.

Table of Content

Displacement Time Graph
Velocity Time Graph
Acceleration time Graph
Things to Remember
Sample Questions

Also check: Potential Energy

Displacement Time Graph

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Displacement is defined by the distance travelled by an object from its original point. In this graph, displacement is a dependent variable and is plotted on the Y-axis while time is an independent variable and is plotted by the x-axis, it is also known as the position time graph. The slope can be illustrated by:

S = displacement / time

= y₂ − y₁ / x₂ − x₁

= Δ d / Δ t

= velocity

Following are three different plots for the displacement time graph:

The first graph represents that an object is stationary for a period of time so that the slope is zero which means that the velocity of the object is zero.
The second graph shows that the velocity of the object is constant and it increases over time so that the slope of the graph remains constant and positive.
The third graph shows that acceleration, velocity, and displacement are constant. Slope of the curve increases along with an increase in time.

Given below are the key takeaways from the displacement time graph:

Velocity of the object is equal to the slope of the curve.
Constant velocity is expressed in a straight line, whereas acceleration is expressed as a curved line.
When an object is at rest, it means that the slope is zero.
The movement in a positive direction indicates a positive slope.
Negative slope implies motion in an opposite direction.

Also check:

Total Internal Reflection	Refraction

Velocity Time Graph

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In the velocity time graph, velocity is the dependent variable and is represented on the Y-axis and time is an independent variable, represented on the X-axis. The slope of the velocity time graph is derived as:

Slope S = Velocity / time

= y₂−y₁ / x₂−x₁

= Δ v / Δ t

= acceleration

We see that the slope of the velocity time graph is the definition of acceleration, so we can say that the slope is equal to acceleration.

We can understand the following points by observing the graphs:

The steep slope of the graph represents a rapid change in velocity.
The shallow slope represents a slow change of velocity.
If the slope of the graph is negative, the acceleration will be negative.
If the slope of the graph is positive, the acceleration will be positive.
The area under the velocity, shows the displacement of the object.

Also check:

Ray Optics	Wave Optics

Acceleration Time Graph

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In the acceleration time graph, acceleration is the dependent variable and is represented on the Y-axis and time is the independent variable and is represented on the X-axis. The slope of the acceleration time graph is as follows:

S = Acceleration / time

= y₂−y₁ / x₂−x₁

= Δ a / Δ t

The slope of the acceleration time graph is also called Jerk. Following are the points that can be concluded from the graph:

If the slope is zero, it means that the motion has constant acceleration.
The area below the graph signifies the change in velocity.

Things to Remember

Graphs are a way of expressing the relationship between two values.
It is a representation of data in a two-dimensional diagram that describes the relationship between dependent and independent variables.
In a displacement time graph, displacement is a dependent variable and is plotted on the Y-axis while time is an independent variable and is plotted by the x-axis.
In the velocity time graph, velocity is the dependent variable and is represented on the Y-axis and time is an independent variable, represented on the X-axis.
In the acceleration time graph, acceleration is the dependent variable and is represented on the Y-axis and time is the independent variable and is represented on the X-axis.

Also Read:

Average Acceleration Formula	Velocity
Vectors	Speed-Time Graphs

Sample Questions

Ques. How do we plot graphs? (2 marks)

Ans Graphs are created from an array of data. Since most of the data collected are quantitative, it is easy to plot tables and charts for easy understanding and use. Graphs also help in the studies of mechanics.

Ques. What are the uses of graphs? (2 marks)

Ans Graphs are the simplest method to illustrate as well as observe the relationship between a set of given data. In two dimensional graphs, there are two-axis, such as the X-axis and the Y-axis.

Ques. Elaborate on the motion on a velocity-time graph. (2 marks)

Ans The velocity-time graphs are used to describe the motion of objects which are moving in a straight line. The direction of the object can change, such as forward to backward, left or right, and up or down. This graph is similar to the velocity-time graphs.

Ques. Under what condition is the equation ‘s = vt’ correct? (2 marks)

Ans The given equation is correct when the object moves with uniform velocity and along a straight line.

Ques. State the characteristics of displacement. (2 marks)

Ans Following are the characteristics of displacement:

i) Displacement is a vector quantity, thus has both magnitude and direction.

ii) Displacement of a given body can be positive, negative, or even zero.

Ques. State whether a carriage moving between two stations can be considered a point object. (2 marks)

Ans The carriage can be treated as a point-sized object because the size of a carriage is very small as compared to the distance between two stations.

Ques. What is the relative velocity of two bodies which have equal velocities? (2 marks)

Ans The relative velocity of two bodies which have equal velocities is given by:

Taking, v_a = v_b = v

Then, v_ab = v_a – v_b

= v – v = 0

Ques. Rahul went on his bike from Pune to Mumbai at a speed of 60km/hr and came back at a speed of 40km/hr. What is the average speed for Rahul’s entire journey? (3 marks)

Ans In the above question it is known to us that:

Speed of the bike when Rahul travelled from Pune to Mumbai is (v₁) = 60 km/hr.

Returning back speed (v₂) = 40 km/hr.

Therefore, average speed = 2v₁v₂ / v₁+v₂

= 2(60)(40) / 60+40

= 48 km/hr

Ques. Two vehicles A and B cross a point P with velocities 10 m/s and 15 m/s. Then they drive with different uniform velocities and A overtakes B with a speed of 25 m/s. Calculate the velocity of B at that point of time? (5 marks)

Ans: Let a_A = a₁ and b_B = a₂

(25)² = (10)²+2a₁S

Or, 2a₁S = 525 .... (1)

25 = 10+a₁t

So, a₁t = 15 .... (2)

Let velocity of B be v. Then

v² = (15)²+2a₂S

Or, 2a₂S = v²-225 .... (3)

v = 15+a₂t

So, a₂t = v – 15 .... (4)

From equations (1) and (2),

2S/t = 525/15 = 35

and from equations (3) and (4),

2S/t = v²-225/v-15

So, 35 = v²-225/v-15

Solving the two equations we get,

v = 20 m/s

Therefore, we can conclude that the velocity of B at that point of time would be 20 m/s.

Also check:

Potentiometer	Ampere
Unit of Voltage	Circuit Diagram

Graphs: Introduction, Comparison

Displacement Time Graph

Velocity Time Graph

Acceleration Time Graph

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

2.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

3.
What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?

4.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

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Graphs: Introduction, Comparison

Displacement Time Graph

Velocity Time Graph

Acceleration Time Graph

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

2.Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

3.What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?

4.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

2.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

3.
What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?

4.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: