Informative line

### Graphs Of Various Parameters Of Motion

Learn graphs of various parameters of motion & difference between accelerating and non-accelerating motion. Practice equation to calculate of acceleration, displacement from v-t graph and conversion from v-t to s-t graph.

# Interpretation of Path and Graph

• Path of a particle is the actual representation of movement of particle whereas graph represents the positions of particle at different time intervals.

For example :

• A car is moving along a straight road in positive y-direction, as shown in figure.
• The car starts its journey from point $$\text{A}$$ with initial velocity '$$v$$' and constant acceleration '$$–a$$'.

• The car moves continuously in positive y-direction until its velocity becomes zero at point $$\text{B}$$, as shown in figure.

• But the acceleration is always in negative y-direction.
• So it begins to move in negative y-direction and reaches again to point $$\text{A}$$, as shown in figure.

### Path of the Car

The path of the car can be shown as follows -

### Position - time Graph

The graph between position and time can be represented as -

• The position - time graph represents that car starts moving from point $$\text{A}$$.
• As the time elapses, it reaches to point $$\text{B}$$.
• Again, it returns back to point $$\text{A}$$.
• Because, position first increases from point $$\text{A}$$ to $$\text{B}$$ then decreases from $$\text{B}$$ to $$\text{A}$$.
• From the given example, it is clear that path and graph both are independent.

#### The position - time graph of a particle in one dimensional motion is shown in the figure, choose the correct option. Assuming the motion of the particle is along y-axis.

A Particle moves in negative y-direction.

B Particle moves in positive y-direction.

C Particle first moves in positive y-direction and then moves in negative y- direction.

D Particle first moves in negative y-direction and then moves in positive y-direction.

×

The position of the particle decreases from S0 to O.

It means particle moves in negative y-direction.

### The position - time graph of a particle in one dimensional motion is shown in the figure, choose the correct option. Assuming the motion of the particle is along y-axis.

A

Particle moves in negative y-direction.

.

B

Particle moves in positive y-direction.

C

Particle first moves in positive y-direction and then moves in negative y- direction.

D

Particle first moves in negative y-direction and then moves in positive y-direction.

Option A is Correct

# Difference between Accelerating and Non-accelerating Motion

## Non-accelerating Motion

• A particle is said to be in non-accelerating motion if the velocity of a particle is constant throughout the motion.
• If particle is moving in positive direction, it will keep on moving in positive direction with the same speed.
• If particle is moving in negative direction, it will keep on moving in negative direction with the same speed.

• Since velocity is constant throughout the motion so acceleration is always zero.

### Accelerating Motion

• A particle starts moving from origin with initial velocity = 0.
• Let acceleration of the particle is in positive direction.
• The acceleration-time graph of the particle will be is that shown in figure.

• As the acceleration of the particle is in positive direction.
• The particle starts moving with increasing speed in positive direction as particle starts with zero initial velocity from origin.
• The velocity-time graph of the particle is that shown in figure.

#### The velocity-time graph of a particle traveling along a straight line is shown in figure. Which one of the following arrangement represents the motion of the particle?

A Non-accelerating motion $$\rightarrow$$ Accelerating motion $$\rightarrow$$ Accelerating motion

B Accelerating motion $$\rightarrow$$ Non-accelerating motion $$\rightarrow$$ Accelerating motion

C Accelerating motion $$\rightarrow$$ Accelerating motion $$\rightarrow$$ Non-accelerating motion

D Non accelerating motion $$\rightarrow$$ Accelerating motion $$\rightarrow$$ Non-accelerating motion

×

Break the motion, wherever graph is changing the directions.

As velocity is constant, the motion is non-accelerating.

As velocity is variable, motion is accelerating.

As velocity is variable, motion is accelerating.

### The velocity-time graph of a particle traveling along a straight line is shown in figure. Which one of the following arrangement represents the motion of the particle?

A

Non-accelerating motion $$\rightarrow$$ Accelerating motion $$\rightarrow$$ Accelerating motion

.

B

Accelerating motion $$\rightarrow$$ Non-accelerating motion $$\rightarrow$$ Accelerating motion

C

Accelerating motion $$\rightarrow$$ Accelerating motion $$\rightarrow$$ Non-accelerating motion

D

Non accelerating motion $$\rightarrow$$ Accelerating motion $$\rightarrow$$ Non-accelerating motion

Option A is Correct

# Comparison of s-t Graph of Two Bodies

• Two particles traveling along a straight line are shown in figure.
• The particles start from origin and travel along positive x-direction simultaneously.
• Both particles are having different velocities.
• As velocity = slope of s-t graph

v1 = slope of graph (1) = $$tan\,\theta_1$$

v2 = slope of graph (2) = $$tan\,\theta_2$$

• As  $$\theta_1>\theta_2$$
• $$tan\,\theta_1>tan\,\theta_2$$

$$v_1>v_2$$

• The velocity of particle (1), v1 > The velocity of particle (2), v2
• To reach certain point A, particle (1) takes $$\Delta t_1$$ time, whereas particle (2) takes $$\Delta t_2$$ time.

$$\Delta t_1<\Delta t_2$$

#### The figure shows s-t graph of two particles traveling along a straight line. Mark the incorrect answer.

A Velocity of particle (1) is more than velocity of particle (2).

B To reach at certain point, particle (2) will take more time.

C To reach at certain point, particle (1) will take more time.

D Both the particles are traveling along positive directions.

×

Slope of s-t curve of (1) > Slope of s-t curve of (2)

Velocity of particle (1) > Velocity of particle (2)

To reach certain point, particle (1) will take less time than (2).

$$\therefore$$ Option (C) is incorrect.

### The figure shows s-t graph of two particles traveling along a straight line. Mark the incorrect answer.

A

Velocity of particle (1) is more than velocity of particle (2).

.

B

To reach at certain point, particle (2) will take more time.

C

To reach at certain point, particle (1) will take more time.

D

Both the particles are traveling along positive directions.

Option C is Correct

# Comparison of v-t Graph of Two Bodies

• Two particles traveling along a straight line with zero initial velocity are shown in figure.
• The direction of motion is positive x-axis.
• They are moving with different accelerations.
• Acceleration = slope of v-t graph
• Acceleration of particle (1) = slope of graph (1) = $$tan\,\theta_1$$
• Acceleration of particle (2) = slope of graph (2) = $$tan\,\theta_2$$

Here,  $$\theta_1>\theta_2$$

$$\therefore\,tan\theta_1>tan\theta_2$$

$$a_1>a_2$$

• Acceleration of particle (1) > Acceleration of particle (2)
• To reach certain value of velocity, particle (2) will take more time.
• To reach v0 velocity, particle (1) takes $$\Delta t_1$$ time and particle (2) takes $$\Delta t_2$$ time.

$$\Delta t_1\,<\,\Delta t_2$$

It means particle (2) takes more time.

## Comparison of Displacement through v-t Graph

• Displacement = Area under v-t graph

• In equal time interval $$\Delta t$$, area under the graph (1) > area under the graph (2)
• Displacement of particle (1) > Displacement of particle (2)

#### The following figure represents v-t graph of particles traveling along a straight line. Choose the correct answer.

A Both will have same displacement in equal time

B Acceleration of particle (2) is less than acceleration of particle (1)

C Both are having equal accelerations

D Acceleration of particle (2) is more than acceleration of particle (1)

×

Acceleration = slope of v-t graph

Slope of particle (1) < Slope of particle (2)

Acceleration of particle (2) > Acceleration of particle (1)

### The following figure represents v-t graph of particles traveling along a straight line. Choose the correct answer.

A

Both will have same displacement in equal time

.

B

Acceleration of particle (2) is less than acceleration of particle (1)

C

Both are having equal accelerations

D

Acceleration of particle (2) is more than acceleration of particle (1)

Option D is Correct

# Calculation of Acceleration from v-t Graph

• To calculate  acceleration from velocity- time graph, consider a particle traveling along a straight line with uniform acceleration.
• The v-t graph of the particle is shown in the figure.
• We know that -

Acceleration = $$\dfrac {\Delta v} {\Delta t}$$$$\dfrac { v_3–v_1} {t_2–0}$$

Now, from the graph,

$$\dfrac {\Delta v} {\Delta t}$$= tan$$\theta$$ $$\implies$$ slope

$$\therefore\,\text{Acceleration = slope of v-t graph}$$

Points to remember :

• To determine slope, $$\theta$$ should always be measured from positive x-axis.
• Also remember that,

$$tan\theta$$ is positive when $$0^0<\theta< 90^0$$

$$tan\theta$$ is negative when  $$90^0<\theta< 180^0$$

$$tan\theta$$ is  $$\infty$$ when $$\theta=90^0$$

• So,
1. If  $$\theta$$ is acute, the slope will be positive.

2. If $$\theta$$ is obtuse, the slope will be negative.

3. If $$\theta$$ is $$90^0$$, the slope will be infinite.

#### The velocity–time graph of a particle, moving along a straight line, is shown in the given figure. Calculate the acceleration of the particle.

A $$3\,m/s^2$$

B $$4\,m/s^2$$

C $$2\,m/s^2$$

D $$5\,m/s^2$$

×

Change in velocity = $$(v_2-v_1)=\Delta v=10\,m/s$$

Change in time = $$(t_2-t_1)=\Delta t=5\,sec$$

Slope of v-t graph,

$$tan\theta=\dfrac{\Delta v}{\Delta t}=2$$

Acceleration of the particle,

$$\dfrac{\Delta v}{\Delta t}=2\,m/s^2$$

### The velocity–time graph of a particle, moving along a straight line, is shown in the given figure. Calculate the acceleration of the particle.

A

$$3\,m/s^2$$

.

B

$$4\,m/s^2$$

C

$$2\,m/s^2$$

D

$$5\,m/s^2$$

Option C is Correct

# Calculation of Displacement from v-t Graph

• The area under v-t curve gives the value of displacement in the given time interval.
• Consider a particle moving along a straight line. The v-t curve of particle is shown in figure.
• Thus,

Displacement in the time interval $$\Delta t$$ = Area under the v-t curve

#### The given figure represents the v-t graph of a particle moving along a straight line. Find displacement of the particle in first 2 sec.

A 25 m

B 30 m

C 35 m

D 26 m

×

Displacement in first 2 seconds = Area covered in first 2 seconds

Area = Area of triangle + Area of rectangle

Area of triangle = $$\dfrac{1}{2}×base × height$$

$$=\dfrac{1}{2}×1.5×20$$

$$=15\,m$$

Area of rectangle = length × Breadth

=0.5 (20)

=10 m

Displacement = 15+10 = 25 m

### The given figure represents the v-t graph of a particle moving along a straight line. Find displacement of the particle in first 2 sec.

A

25 m

.

B

30 m

C

35 m

D

26 m

Option A is Correct

# Conversion from v-t to s-t Graph

• v-t graph of a particle traveling along a straight line is shown in figure.
• Particle starts from origin and is traveling along x-axis.
• Velocity = slope of s-t graph

For $$\Delta t_1$$ time :

• Velocity of particle is positive, slope of s-t graph during $$\Delta t_1$$ time interval is also positive.
• Position will always increase.

For $$\Delta t_2$$ time interval :

• Velocity < 0
• Slope of s-t graph < 0
• Position will decrease.

For $$\Delta t_3$$ time interval :

Velocity = 0,

$$\therefore$$ Slope of s-t graph = 0

• The particle will stop and will remain at the same position.

#### The velocity-time graph of a particle traveling along a straight line is shown in figure. Mark the correct position-time graph corresponding to it. Assume particle starts from origin.

A

B

C

D

×

Break the motion, wherever graph is changing directions.

Velocity in 0 to 1 sec. time interval = 5 m/s

Slope of s-t curve in 0-1 sec time interval will be positive and equal to 5.

Position after $$\Delta t_1=1\,sec$$

$$\Rightarrow\,\,\,s=v_1(\Delta t_1)$$

$$s=5×1=5\,m$$

Velocity in 1 – 2 sec time interval = –7 m/s

Slope of s-t curve in 1– 2 sec time interval will be negative and equal to 7.

s-t curve till $$\Delta t_2$$ time interval

Position after 2 sec

$$s=s_o+v_2\,\Delta t_2$$

$$s=5+(–7)(1)$$

$$s=\,–2\,m$$

Velocity in 2 – 3 sec time interval = 5 m/s

Slope of s-t curve in 2 - 3 sec time interval will be positive and equal to 5.

Position after

$$\Delta t_3 =3\; \text {sec}$$

$$s=s_o+v_3\,\Delta t_3$$

$$s=\;–2+5(1)$$

$$s=3\,m$$

### The velocity-time graph of a particle traveling along a straight line is shown in figure. Mark the correct position-time graph corresponding to it. Assume particle starts from origin.

A
B
C
D

Option A is Correct

# Conversion from a-t to v-t Graph

• A particle is moving along x-axis with zero initial velocity.
• The acceleration-time graph of the particle is shown in figure.
• Acceleration = slope of v-t graph

In $$\Delta t_1$$ time interval

Acceleration, a > 0

slope of v-t curve > 0

Velocity will increase.

For $$\Delta t_2$$  time interval

Acceleration < 0

slope of v-t graph < 0

Velocity of a particle will decrease and it will increase in negative direction.

For $$\Delta t_3$$  time interval

Acceleration = 0

slope of v-t graph = 0

$$\therefore$$ Velocity in $$\Delta t_3$$ time interval is constant.

#### The a-t graph of a particle traveling along a straight line is shown in the figure. Choose the corresponding v-t graph of the particle. Given Initial velocity of the particle is 2 m/sec.

A

B

C

D

×

Break the motion, wherever graph is changing directions.

Acceleration in 0 – 1 sec time interval = 10 m/s2

Slope of v-t curve in 0-1 sec will be positive and equal to 10.

v- t curve for $$\Delta t_1$$

$$v_1=v_i+a_1\,\Delta t_1$$

$$v_1=2+10(1)$$

$$v_1=12\,m/s$$

(Initial velocity, $$v_i$$ = 2 m/s)

Acceleration in 1 - 2 sec time interval = –12 m/s2

Slope of v-t graph 1 - 2 sec will be negative and equal to 10.

v-t curve till $$\Delta t_2$$

$$v_2=v_1+a_2\,\Delta t_2$$

$$v_1=12\,m/s$$

$$v_2=12–12(1)$$

$$v_2=0$$

Acceleration in 2 - 3 sec time interval = 2 m/s2

slope of v-t graph in 2 - 3 sec time interval will be positive and equal to 2.

v-t curve till $$\Delta t_3$$

$$v_3=v_2+a_2(\Delta t_2)$$

$$=0+2(1)$$

$$v_3=2\,m/s$$

### The a-t graph of a particle traveling along a straight line is shown in the figure. Choose the corresponding v-t graph of the particle. Given Initial velocity of the particle is 2 m/sec.

A
B
C
D

Option C is Correct

# Conversion of s-t Graph to v-t Graph

• Consider a s-t graph of a particle moving along a straight line. The graph is shown in figure.

• To convert s-t graph to v-t graph, following relation is used,

$$v=\dfrac{ds}{dt}$$= slope of s-t graph

• We will find the values of velocities for each given time interval by calculating the slope of s-t graph.
• Velocity of the particle in $$\Delta t_1=t_1–0$$, time interval = slope of s-t graph

$$v=\dfrac{ds}{dt}=tan\,\theta$$

• Velocity of the particle in $$\Delta t_2=t_2–t_1$$, time interval = slope of s-t graph = 0
• Thus, v-t graph can be given as -

#### The s-t graph of a particle moving along a straight line is shown in figure. Choose the correct v-t graph for the particle.

A

B

C

D

×

Velocity during first 3 seconds = slope of s-t graph

$$\dfrac{ds}{dt}$$$$\dfrac{10–10}{3–0}=\dfrac{0}{3}=0$$

Velocity during next 2 seconds = slope of s-t graph in this interval

$$v=\dfrac{0–10}{5–3}=–5\,m/s$$

Velocity-time graph of the particle -

### The s-t graph of a particle moving along a straight line is shown in figure. Choose the correct v-t graph for the particle.

A
B
C
D

Option D is Correct

# Conversion of v-t to a-t Graph

• A particle is moving along a straight line, the v-t graph of the particle is shown in figure.

• To convert v-t graph to a-t graph.
• Find accelerations for different time intervals.
• Acceleration of the particle in $$\Delta t_1$$ time interval,

a1 = slope of v-t graph = $$tan\,\theta_1$$

• Acceleration of the particle in $$\Delta t_2$$ time interval,

a2 = slope of v-t graph $$=tan\theta_2$$

• Acceleration of the particle in $$\Delta t_3$$ time interval,

a3 = slope of v-t graph $$=tan\theta$$

So, the a-t graph of the particle will be

#### Velocity - time graph of a particle moving along a straight line is shown in the following figure. Choose the correct a-t graph corresponding to the given graph.

A

B

C

D

×

Lets break the motion where graph is changing for different time interval.

Acceleration during the first 5 sec

$$a_1=tan\,0^o=0$$

As the angle made from positive x-axis is 0o.

Acceleration during the next 5 sec (5 to 10 sec)

$$a_2=tan\,\theta$$

$$a_2=tan\,(\pi –\phi)$$

$$a_2= \,–\,tan\, \phi$$

From  $$\Delta \,ABC$$,

$$tan\,\phi=\dfrac{20}{5}=4$$  ;  $$tan\,\theta =\, –4$$

$$\Rightarrow\,\,a_2=\;–4\,m/s^2$$

Acceleration during the next 5 sec (10 to 15 sec)

$$a_3=tan\,\theta_1$$

From  $$\Delta \,CDE$$,

$$tan\theta_1=\dfrac{14}{5}$$

$$a_3=2.8\;m/s^2$$

So, the a-t curve of the particle will be

### Velocity - time graph of a particle moving along a straight line is shown in the following figure. Choose the correct a-t graph corresponding to the given graph.

A
B
C
D

Option B is Correct