Informative line

### Uniform Velocity

Learn uniform motion definition with calculate of average velocity and speed including distance & displacement.

# Identification of Uniform Motion

• A body is said to be in uniform motion if its direction of motion and speed remains unchanged during a given time interval.
• For one Dimensional motion, a particle is said to be in uniform motion if it keeps on moving either in positive or negative direction with same speed,i.e., it covers equal distance in equal interval of time.

Note :- For speed to be constant, equal distance should be covered in equal interval of time.

• Check whether the direction is constant throughout the motion.
• Check whether the speed is constant throughout the motion.
• If both are constant then motion is uniform.
• If any of the two is not constant then motion is non-uniform.

#### The following data represents positions of a particle at different times.Which type of motion does the particle possess?

A Uniform motion

B Non-uniform motion

C Can't be determined

D

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Controller::render() - CORE/Cake/Controller/Controller.php, line 952
Dispatcher::_invoke() - CORE/Cake/Routing/Dispatcher.php, line 192
Dispatcher::dispatch() - CORE/Cake/Routing/Dispatcher.php, line 160
[main] - APP/webroot/index.php, line 108

×

Direction of motion is in positive  $$x$$- direction and it is same throughout the motion.

Speed of the particle is constant throughout the motion and is equal to 5 m/s.

Since, the direction of motion and the speed, are constant throughout the motion does motion is uniform.

### The following data represents positions of a particle at different times.Which type of motion does the particle possess?

A

Uniform motion

.

B

Non-uniform motion

C

Can't be determined

Notice (8): Undefined offset: 3 [APP/View/Elements/theorypage.ctp, line 285]"> D
Notice (8): Undefined offset: 3 [APP/View/Elements/theorypage.ctp, line 289]Code Context

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$str = '<h1><strong>Identification of Uniform Motion</strong>&nbsp;</h1> <ul> <li>A body is&nbsp;said to be in uniform motion if&nbsp;its direction of motion and speed remains unchanged during a given time interval.&nbsp;</li> <li>For one Dimensional motion, a particle&nbsp;is said to be in uniform motion if it keeps on moving either in positive or negative direction with same speed,i.e., it covers equal distance in equal interval of time.&nbsp;</li> </ul> <p><strong>Note :-</strong>&nbsp;For speed to be constant, equal distance&nbsp;should be covered&nbsp;in equal interval of time. &nbsp;</p> <ul> <li>Check whether the direction is constant throughout the motion.</li> <li>Check whether the speed is constant throughout the motion.&nbsp;</li> <li>If both are constant then motion is uniform.</li> <li>If any of the two is not constant then motion is non-uniform.</li> </ul> '$pos = (int) 0
$pos2 = (int) 59$header = '<h1><strong>Identification of Uniform Motion</strong>&nbsp;'
$m = (int) 3$correctOption = 'A'include - APP/View/Elements/theorypage.ctp, line 289
View::_evaluate() - CORE/Cake/View/View.php, line 935
View::_render() - CORE/Cake/View/View.php, line 897
View::_renderElement() - CORE/Cake/View/View.php, line 1195
View::element() - CORE/Cake/View/View.php, line 412
include - APP/View/Students/subject.ctp, line 10
View::_evaluate() - CORE/Cake/View/View.php, line 935
View::_render() - CORE/Cake/View/View.php, line 897
View::render() - CORE/Cake/View/View.php, line 466
Controller::render() - CORE/Cake/Controller/Controller.php, line 952
Dispatcher::_invoke() - CORE/Cake/Routing/Dispatcher.php, line 192
Dispatcher::dispatch() - CORE/Cake/Routing/Dispatcher.php, line 160
[main] - APP/webroot/index.php, line 108
Notice (8): Undefined offset: 3 [APP/View/Elements/theorypage.ctp, line 296]Code Context                                            <?php

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Option A is Correct

# Finding Position of a Particle After Given Time

• Consider a particle is moving along $$x$$- axis with uniform velocity $$v_x$$.
• Now, let its initial position be $$x_i$$ at time $$t= t_1$$ and its final position be $$x_f$$ at time $$t=t_2$$.

Now,    $$x_f = v_x.t_2\;\;\;\;\;\;\;\;— (1)$$

$$x_i = v_x. t_1 \;\;\;\;\;\;\;— (2)$$

Subtracts (2) from (1) we get,

$$x_f-x_i = v _x (t_2 - t_1 )$$

or,  $$x_f = x_i + v_x (t_2 - t_1)$$

or,  $$x_f = x_i + v_x \; \Delta t$$

Note- While solving the problems we can use time interval $$(\Delta t)$$ directly. It is not necessary to have initial and final values of time.

#### A particle starts moving from $$x = - 2\; m$$ with a constant velocity of $$- 3 \; m/s$$. Find the position of the particle after 2 sec.

A – 8 m

B – 10 m

C 10 m

D 5 m

×

Initial position at time $$t_1$$

$$x_i = - 2\; m$$

Velocity $$= - 3 \; m/s$$

Final position at  time $$t_2$$

$$x_f = x_i + v_x (t_2 - t_1)$$

$$x_f = - 2 + (-3) (2)$$

$$x_f = - 8 \; m$$

### A particle starts moving from $$x = - 2\; m$$ with a constant velocity of $$- 3 \; m/s$$. Find the position of the particle after 2 sec.

A

– 8 m

.

B

– 10 m

C

10 m

D

5 m

Option A is Correct

# Definition of Average Velocity

• If displacement of a particle is $$\Delta x$$ in a time interval $$\Delta t$$ then, its average velocity during this time interval will be

$$v_{avg} = \dfrac{\Delta x}{\Delta t}$$

$$v_{avg} = \dfrac{\text{Total displacement}}{\text{Total time}}$$

• If a particle moving along a straight line, is at position $$x_1$$ at time $$t = t_1$$ and at position $$x_2$$ at time $$t= t_2$$ then, average velocity of the particle is,

$$v_{avg} = \dfrac{x_2 - x_1}{t_2 - t_1}$$

• Initial position at $$t = t_1$$ is $$x_1$$.
• Final position at $$t = t_2$$ is $$x_2$$.
• Time interval = $$(t_2 - t_1) = \Delta t$$

Average velocity,

$$v_{avg} = \dfrac{x_2 - x_1}{t_2 - t_1}$$

#### The position of a particle moving along a straight line is $$3\; m$$ at time  $$t= 0\,sec$$ and $$- 15 \; m$$ at time $$t = 3 \;sec$$ . Calculate average velocity of the particle during the given time interval.

A – 6 m/s

B – 8 m/s

C – 10 m/s

D – 15 m/s

×

Initial position at $$t_i= 0\; sec$$

$$x_i = 3 \; m$$

Final position at $$t_f = 3\; sec$$

$$x_f = - 15 \; m$$

Time interval $$\Delta t = t_f - t_i = 3 \; sec$$

Average velocity,

$$v_{avg} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_f-x_i}{t_f - t_i}$$

$$v_{avg} = \dfrac{-15 - 3}{3} = -6 \; m/s$$

### The position of a particle moving along a straight line is $$3\; m$$ at time  $$t= 0\,sec$$ and $$- 15 \; m$$ at time $$t = 3 \;sec$$ . Calculate average velocity of the particle during the given time interval.

A

– 6 m/s

.

B

– 8 m/s

C

– 10 m/s

D

– 15 m/s

Option A is Correct

# Average Velocity:Different period with Different Speeds

• Consider a particle is moving along a straight line with velocity $$v_1$$ for the first $$t_1$$ sec and with velocity $$v_2$$ for the next $$t_2$$ sec.
•  Velocity for first $$t_1$$ sec$$= v_1$$
• Velocity for next $$t_2$$ sec$$= v_2$$
• Total time interval $$=t_1+t_2$$
• Total displacement $$= \; v_1\;t_1 + v_2 \; t_2$$
• Average velocity, $$v_{avg } = \dfrac{v_1 \; t_1 + v_2\;t_2}{t_1 + t_2}$$

#### A particle is moving along a straight line with a velocity of $$5 \; m/s$$ for the first $$2 \; sec$$ and $$- 8 \; m/s$$ for the next $$5\; sec$$ . Calculate average velocity of the particle.

A 5 m/s

B 6 m/s

C – 4.28 m/s

D 7 m/s

×

Velocity for first $$2 \;sec$$

$$= 5\; m/s$$

Velocity for next $$5\ sec$$

$$= - 8\; m/s$$

Total time interval

$$= t_ 1 +t_2 = 7 \; sec$$

Total displacement $$(\Delta x)$$

$$=v_1 \; t_1 + v_2 \; t_2$$

$$=\; 5\; (2) + (-8) \; (5)$$

$$= - 30 \; m$$

Average velocity $$(v_{avg})$$

$$= \dfrac{\Delta x}{\Delta t} = \dfrac{-30}{7}$$

$$= - 4.28 \; m/s$$

### A particle is moving along a straight line with a velocity of $$5 \; m/s$$ for the first $$2 \; sec$$ and $$- 8 \; m/s$$ for the next $$5\; sec$$ . Calculate average velocity of the particle.

A

5 m/s

.

B

6 m/s

C

– 4.28 m/s

D

7 m/s

Option C is Correct

# Calculation of Displacement

• Consider a particle is moving along a straight line with a velocity $$v_1$$ for first $$t_1\; sec$$ and with velocity $$v_2$$ for the next $$t_2 \; sec$$ .
• So,

Displacement $$\Delta x_1$$ in first $$t_1 \; sec = v_1 \; t_1$$

Displacement $$\Delta x_2$$ in next $$t_2 \; sec = v_2 \; t_2$$

• So,

Total displacement during $$(t_1 + t_2 )$$ sec

$$= v_1 \; t_1 + v_2 \; t_2$$

#### A particle is moving along a straight line with a velocity of $$3 \; m/s$$ for first $$5 \; sec$$ and $$- 8\; m/s$$ for the next $$5 \; sec$$. Calculate total displacement of the particle.

A – 30 m

B – 25 m

C – 35 m

D 40 m

×

Displacement in first $$5 \; sec$$

$$\Delta x_1 = 5 × 3$$

$$\Delta x_1 = 15 \; m$$

Displacement in next $$5 \; sec$$

$$\Delta x_2 = 5 ×(-8)$$

$$\Delta x_2 = - 40 \; m$$

Total displacement $$(\Delta x)$$

$$\Delta x = \Delta x_1 + \Delta x_2$$

$$\Delta x = 15 - 40$$

$$\Delta x = - 25 \; m$$

### A particle is moving along a straight line with a velocity of $$3 \; m/s$$ for first $$5 \; sec$$ and $$- 8\; m/s$$ for the next $$5 \; sec$$. Calculate total displacement of the particle.

A

– 30 m

.

B

– 25 m

C

– 35 m

D

40 m

Option B is Correct

# Calculation of Time of Journey

• Consider a particle which covers a distance s1 with a uniform speed v1 and another distance s2 with a uniform speed v2.
• Distance covered with speed v1 =  S1
• Time taken to cover s1

$$t_1 = \dfrac{s_1}{v_1}$$

• distance covered with speed v2 =  S2
• Time taken to cover s2

$$t_2 = \dfrac{s_2}{v_2}$$

• Total time of journey $$= t_1 + t_2$$

$$=\dfrac{s_1}{v_1} + \dfrac{s_2}{v_2}$$

#### A particle covers a distance of $$25 \; m$$ with a speed of $$5 \; m/s$$ and then covers $$24 \; m$$ with a speed of $$8 \; m/s$$ . Calculate the total time of journey.

A 8 sec

B 10 sec

C 9 sec

D 1 sec

×

Distance covered with speed $$5\; m/s$$

$$= 25 \; m$$

Time taken to cover $$25 \; m$$

$$= \dfrac{25}{5} = 5 \; sec$$

Distance covered with speed $$8 \; m/s$$

$$= 24 \; m$$

Time taken to cover $$24 \; m$$

$$= \dfrac{24}{8} = 3 \; sec$$

Total time of journey = $$5 +3 = 8 \; sec$$

### A particle covers a distance of $$25 \; m$$ with a speed of $$5 \; m/s$$ and then covers $$24 \; m$$ with a speed of $$8 \; m/s$$ . Calculate the total time of journey.

A

8 sec

.

B

10 sec

C

9 sec

D

1 sec

Option A is Correct

# Advanced Problems on Average Speed

• Consider a particle which covers a distance s1 with a speed v1 and distance s2 with a speed v2
• Time taken to cover distance s1,

$$t_1=\dfrac{s_1}{v_1}$$

• Time taken to cover distance s2,

$$t_2=\dfrac{s_2}{v_2}$$

• $$\text{Average speed }=\dfrac{\text{Total distance }}{\text{Total time}}$$

$$= \dfrac{s_1 + s_2}{\dfrac{s_1}{v_1} + \dfrac{s_2}{v_2}}$$

#### Tara covers one third of her journey with a speed of $$25 \; m/s$$ and remaining two third with a speed of $$40 \; m/s$$. Calculate average speed of Tara.

A $$\dfrac{20}{3}\; m/s$$

B $$90\; m/s$$

C $$\dfrac{100}{3}\; m/s$$

D $$\dfrac{60}{3}\; m/s$$

×

Let the total distance covered by Tara is $$x\; m$$ .

Time taken to cover one third of journey

$$= t_1 = \dfrac{\dfrac{x}{3}}{25} = \dfrac{x}{75}\; sec$$

Time taken to cover two third of her journey

$$= t_2 = \dfrac{\dfrac{2}{3}\; x}{40} = \dfrac{x}{60} \; sec$$

$$\text{Average speed of Tara}=\dfrac{\text{Total distance }}{\text{Total time }}$$

$$= \dfrac{x}{\dfrac{x}{75} + \dfrac{x}{60}} = \dfrac{100}{3} \; m/s$$

### Tara covers one third of her journey with a speed of $$25 \; m/s$$ and remaining two third with a speed of $$40 \; m/s$$. Calculate average speed of Tara.

A

$$\dfrac{20}{3}\; m/s$$

.

B

$$90\; m/s$$

C

$$\dfrac{100}{3}\; m/s$$

D

$$\dfrac{60}{3}\; m/s$$

Option C is Correct

# Average Velocity:Different Distance with Different Speed

• Consider a particle which covers a distance s1 with a speed v1 and distance s2 with a speed v2
• Time taken to cover distance s1,

$$t_1=\dfrac{s_1}{v_1}$$

• Time taken to cover distance s2,

$$t_2=\dfrac{s_2}{v_2}$$

• $$\text{Average speed }=\dfrac{\text{Total distance }}{\text{Total time}}$$

$$= \dfrac{s_1 + s_2}{\dfrac{s_1}{v_1} + \dfrac{s_2}{v_2}}$$

#### Tina drives a car with a speed of $$40 \; m/s$$ for first $$200\; m$$ and $$25 \; m/s$$ for the next $$250 \; m$$. Calculate average speed of the car.

A 20 m/s

B 30 m/s

C 40 m/s

D 50 m/s

×

Given,

$$s_1 = 200 \; m, \;\;\;s_2 = 250 \; m$$

Time taken to cover first $$200 \; m$$

$$t_1 = \dfrac{200}{40} = 5\; sec$$

Time taken to cover next $$250 \; m$$

$$t_2 = \dfrac{250}{25} = 10 \; sec$$

$$\text{Average speed } = \dfrac{\text{Total distance }}{\text{Total time}} = \dfrac{s_1 + s_2 }{t_1 +t_2}$$

Average speed = $$\dfrac{250 + 200}{ 15}$$

= $$30 \; m/s$$

### Tina drives a car with a speed of $$40 \; m/s$$ for first $$200\; m$$ and $$25 \; m/s$$ for the next $$250 \; m$$. Calculate average speed of the car.

A

20 m/s

.

B

30 m/s

C

40 m/s

D

50 m/s

Option B is Correct