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.

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.

Illustration Questions

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 = \((t_2-t_1)=\Delta t=5\,sec\)

image

Slope of v-t graph,

 \(tan\theta=\dfrac{\Delta v}{\Delta t}=2\)

image

Acceleration of the particle,

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

image

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

image
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

Illustration Questions

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 = \(\dfrac{1}{2}×base × height\)

\(=\dfrac{1}{2}×1.5×20\)

\(=15\,m\)

image

Area of rectangle = length × Breadth

 =0.5 (20)

=10 m

image

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.

image
A

25 m

.

B

30 m

C

35 m

D

26 m

Option A 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 -

Illustration Questions

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

\(\dfrac{ds}{dt}\)\(\dfrac{10–10}{3–0}=\dfrac{0}{3}=0\)

image

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

\(v=\dfrac{0–10}{5–3}=–5\,m/s\)

image

Velocity-time graph of the particle -

image

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.

image
A image
B image
C image
D image

Option D is Correct

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.

Illustration Questions

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.

image

It means particle moves in negative y-direction.

image

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.

image
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

  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

Illustration Questions

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.

image

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

\(a_2=tan\,\theta\)

\(a_2=tan\,(\pi –\phi)\)

\(a_2= \,–\,tan\, \phi\)

From  \(\Delta \,ABC\),

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

\(\Rightarrow\,\,a_2=\;–4\,m/s^2 \)

image

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

\(a_3=tan\,\theta_1\)

From  \(\Delta \,CDE\),

\(tan\theta_1=\dfrac{14}{5}\)

\(a_3=2.8\;m/s^2 \)

image

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

image

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. 

image
A image
B image
C image
D image

Option B 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.

Illustration Questions

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.

image

As velocity is variable, motion is accelerating.

image

As velocity is variable, motion is accelerating.

image

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?

image
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.

\(\Delta t_1<\Delta t_2\) 

Illustration Questions

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)

image

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

image

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

image

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

image

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

image
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

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.

Illustration Questions

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

image

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

Position after \(\Delta t_1=1\,sec\)

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

\(s=5×1=5\,m\)

image

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

image

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

\(s=s_o+v_2\,\Delta t_2\)

\(s=5+(–7)(1)\)

\(s=\,–2\,m\)

image

 

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

image

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

Position after

\(\Delta t_3 =3\; \text {sec}\)

\(s=s_o+v_3\,\Delta t_3\)

\(s=\;–2+5(1)\)

\(s=3\,m\)

image

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.

image
A image
B image
C image
D image

Option A 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)

Illustration Questions

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

image

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

image

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

image

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

image
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

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.

Illustration Questions

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

image

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)

image

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

image

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

v-t curve till \(\Delta t_2\)

\(v_2=v_1+a_2\,\Delta t_2\)

\(v_1=12\,m/s\)

\(v_2=12–12(1)\)

\(v_2=0\)

image

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

image

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

v-t curve till \(\Delta t_3\)

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

\(=0+2(1)\)

\(v_3=2\,m/s\)

image

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.

image
A image
B image
C image
D image

Option C is Correct

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