Distance and displacement may or may not be equal in magnitude.

Explanation:

In certain situations, the total distance travelled by an object and its displacement can be the same. In most cases, the distance travelled is greater than the displacement because the object may move along a curved path, making the actual path longer than the shortest distance between two points. Therefore, displacement is always less than or equal to the distance travelled.

Acceleration of the bullet is larger in the gun with the shorter barrel.

Reason:

Because the bullet accelerates over a shorter distance and, according to the equation:

vf² − vi² = 2as

 a greater acceleration is required to achieve a higher velocity in a shorter distance.

Yes, it is possible for the instantaneous velocity to be negative at certain points even if the average velocity is positive.

Justification:

The average velocity depends on the total displacement, but instantaneous velocity can be negative if the object is moving in the opposite direction during certain intervals of the trip.

The graph correctly represents the motion.

Explanation:

In graph C, the velocity is maximum at zero time. While going up, its acceleration due to gravity is negative. At half time of flight, its velocity becomes zero. It comes back downward and returns to the ground with velocity equal to the initial velocity in magnitude.

Yes, it is possible that the velocity of an object is zero at an instant of time.

Example:

A ball is thrown vertically upward. At the highest point , the velocity is zero but acceleration is still 9.8 metre per second downward due to gravity.

A vector is graphically represented by a straight line with an arrow at one end.

Magnitude:

The length of the line represents the magnitude of the vector according to a suitable scale.

Direction:

The direction of the arrow indicates the direction of the vector.

Reference Axes:

Vectors are often represented with respect to two mutually perpendicular reference axes.

These axes are typically:

(i) x-axis : Horizontal axis
(ii) y-axis : Vertical axis

The point where these axes meet is called the origin, usually denoted as O.

Representation of Axes:

To represent the direction, two mutually perpendicular lines are required. We can draw one line to represent east-west direction and other line to represent north-south direction.

The direction of the vector can be given with respect to these lines.

The direction of the vector is also described with respect to the reference axes, often using an angle (θ).

The angle (θ) is measured counterclockwise from the positive direction of the x-axis.

(i) Rest:

Definition:

If a body does not change its position with respect to its surroundings, it is said to be at rest.

Example:

If a motorcyclist is standing on the road and observer will notice that the motorcyclist is not changing his position relative to nearby object such as a building, tree, or electric pole.

In this case, the motorcyclist is considered to be at rest.

(ii) Motion:

Definition:

If a body continuously changes its position with respect to its surroundings,

It is said to be in motion.

Example:

When the motorcyclist starts driving the observer will see that motorcyclist is moving relative to the nearby buildings, tree or electric pole. This means the motorcyclist is in motion.

2. Speed and velocity:

Speed:

Speed is the rate at which a body covers a distance in given time. It is a scalar quantity, which means it has magnitude but no direction.

Formula:

The speed v is calculated as:

$speed=TimeDistance$ $v=ts$

where:

• s = distance covered
• t = time taken

Unit:

SI unit of speed is meter per second or kilometers per hour.

Velocity:

The net displacement of a body in unit time is called velocity.

Formula:

Velocity (v) = Displacement/Time

where;

Displacement =d  and Time =t

where displacement refers to the shortest straight-line between the starting and ending points, including directions.

Unit:

SI unit of velocity is also metre per second.

Types of Motions:
                      In daily life, we observe the main types of motion.
i) Translatory Motion

ii) Rotatory Motion

iii) Vibratory Motion

i- Translatory Motion :
                   "If every particle of a body moves uniformly in the same direction, the motion is called translatory mation.
Examples:
                    The motion of a train or a car.

Types of Translatory Motion:
a) Linear Motion:
                      When a Car moves along a straight line, its motion is called Linear motion.
Example:

                     A freely falling  object.

b) Circular Motion:
                     When a body moves along a  circular path, its motion called circular motion.
Examples:
                  i). A ball tied to a string and whirled.
                  ii). A Ferris wheel.
c) Random Motion:
                  When a body moves along an an irregular or unpredictable path,its motion is called random motion.

Example :
                  The movement of a butterfly.
ii- Rolatory Motion:

If every point of a body moves around a fixed axis, the motion is called rotatory motion.
Examples:
               i- The blades of an electric fan.
               ii- The drum of a washing machine dryer.
               iii- A spinning top.
iii- Vibradory Motion:
                      When a body repeats its  to and fro motion about a fixed position, the motion is  called vibratory motion.
Examples:
               i- A Swing in a park.
               ii- The vibration of a guitar string.

                                       |                                                                          

Distance-Time graph:
The gradient of a  distance-time graph represent the rate of  change of distance with respect to time. This  is equivalent to average speed of the body  during the given time interval. A steeper gradient  indicates a higher speed while a flater gradient indicates a lower speed.

Calculation of Gradient (Slope):
The gradient (slope) is given by:
Gradient = change in distance(s)/change in time(t) = s2 - s1/t2 - t1 =s/t
           

Relate to Average Speed:
From the formula;
Gradient = s/t = v
          

where v is the average speed of the body.

Graphical Interpretation:

The gradient equals tanθ  where θ is the angle formed by the graph  with the horizontal axis.
Hence, Gradient = tanθ = v
In conclusion, the gradient of distance time  graph is equal to the average speed of the body.

Gradient of Speed–Time Graph:

The gradient (slope) of a speed–time graph indicates the average acceleration of an object. It represents the rate of change of speed with respect to time.

Formula for Gradient:

The gradient is calculated as:

Gradient = change in speed (ΔV)/change in Time (Δt) =V₂-V₁/t₂-t₁ = a

where,

(i) V₂ and V₁ are speeds at times t₂ and t₁ respectively.

(ii) a is the average acceleration.

Case 1: Motion with Constant Acceleration

A positive Gradient indicates constant acceleration.

Case 2: Motion with constant speed

A zero gradient indicates motion aat constant speed with zero acceleration.

Area under Speed-Time Graph

The area under a speed-time graph represents the distance travelled by an object during a given time interval. It is calculated by finding the geometric area under the curve up to the time axis.

Case 1: Motion with Constant Speed

When an object moves with a constant speed, the speed-time graph is a horizontal line.

Formula:

$Distance=v\times t$

Geometric Interpretation:

(i).The area under graph forms a rectangle.

(ii). Area of rectangle = Base × Height

$Area=t\times v$

(iii). This area is equal to the distance travelled.

Case 2: Motion with Uniform Acceleration

When an object moves with uniform acceleration, its speed-time graph is a straight line sloping upward from O to V.

Formula:

The average speed during motion is:

Vavg​ =( 0+v)/2 = v/2

Distance Travelled is 

Distance = ​Vavg​ x t = v/2 x t =( 1/2)v x t 

Geometric Interpretation:

The area under the graph forms a right-angled triangle.

Area of Triangle = (1/2)Base x Perpendicular

Area = (1/2)t x v

​This area represents the distance travelled.

1. Equations of Motion Under Gravity:

When solving problems of motion under gravity, the following equations of motion are used:

(i) Final Velocity Equation:

Vf = Vi + at

Where:

Vi = initial velocity
Vf = final velocity
a  = acceleration
t  = time

(ii) Displacement Equation:

S = Vit + 1/2 at²

Where:

S = the displacement or distance covered.

(iii) Velocity–Displacement Equation:

2aS = Vf² − Vi²

2. Assumptions for Applying These Equations:

(i) Straight Line Motion:

Motion is considered along a straight line, simplifying vector analysis.

ii. Vector Magnitudes:

Only the magnitudes of vector quantities are used for calculations.

iii. Uniform Acceleration:

The acceleration (a) is assumed to be constant throughout the motion.

iv. Sign Convention:

• The direction of initial velocity is taken positive.

• Quantities in the same direction as the initial velocity are positive.

• Quantities opposite to the initial velocity are negative (e.g., gravitational acceleration acting downward if the initial velocity is upward).

3. Conclusion:

These equations and assumptions help solve problems for objects in free fall or projected upward under the influence of gravity. Accurate application of these principles ensures correct results for motion analysis.

(a) Solution:

Given data:

V = 400 m s⁻¹
θ = 60°

In scale:

100 m s⁻¹ = 1 cm

So,

400 m s⁻¹ = 4 cm

(b) Solution:

Given data:

Force, F = 50 N
θ = 120°

In scale:

10 N = 1 cm

So,

50 N = 5 cm

Given data:

Average speed (v) = 72km/h
Distance (s) = 360km
To Find:

Time (t) = ?

Solution:
Average speed = Total distance covered / Total time taken
Vav =s/t
By putting values
72 = 360/t
t=360/72
t=5hours

Given data:

Initial velocity (vi) = 0m/s
Final velocity (vf)  = 90km/h                             : 1km = 1000m

                            =(90 x1000)/3600                 :  1h = 60 x 60
                            =25m/s                                         = 3600 s        

        Time (t)       = 50s

To find:

Average acceleration (a_avg) = ?

Solution:

By using first equation of motion 

             a_avg = (25 - 0)/50

                    = 25/50

                    = 0.5 m/s²

Given Data:

Initial velocity (vi) = 5ms⁻¹
       Acceleration (a) = 1.5ms⁻²
       Time (t) = 5s

To find:

Final velocity (vf) = ?
Solution:   Using 1st equation of motion
       vf = vi + at
       By putting values
       vf = 5 + (1.5)(5)
       vf = 5 + 7.5
       vf = 12.5 ms⁻¹

Given data:
Initial velocity (vi) = 18kmh⁻¹
                      = 18 x 1000
                           3600
                      = 180
                           36
                     = 5ms⁻¹

Acceleration (a) = 2ms⁻²
Time (t) = 10sec

To find:
Distance(s) = ?

Solution:
Using 2nd equation of motion
s = vit + 1/2 at²

By putting values
s = (5)(10) + (1/2)(2)(10)²
s = 50 + 100
s = 150m

Given Data:

Initial velocity (vi) = 54kmh⁻¹
                      = 54 x 1000
                           60 x 60
                      = 540
                           6
                      = 15ms⁻¹

Final velocity (vf) = 0ms⁻¹
Distance (s) = 25m

To find:
acceleration (a) = ?

Solution:
Using 3rd equation of motion
2as = vf² - vi²
a = vf² - vi²
         2s

By putting values
a = (0)² - (15)²
        2(25)

a = -225
        50

a = -4.5ms⁻²

Given Data:
Initial velocity (vi) = 0ms⁻¹
height = s = 45m
g = 10ms⁻²

To find:
vf = ?
t = ?

Solution:
Using 3rd equation of motion
2gs = vf² - vi²

Putting values
2(10)(45) = vf² - 0
900 = vf²
√vf² = √900
vf = 30ms⁻¹

For time we use first equation of motion
vf = vi + gt
vf - vi = gt

(vf - vi)/g  = t

 by Putting values
30 - 0 = t
  10
t = 3s

The stone will take 3s to reach the ground.

Given Data :
v = 20ms⁻¹
v₂ = 4ms⁻¹

S₁ = 10km = 10 x 1000 = 10000m
S₂ = 0.8km = 0.8 x 1000 = 800m

average velocity (vav) = ?
average velocity = Total displacement       ....eq (1)
                                    Total time

First consider:
t₁ = S₁
       V₁
t₁ = 10,000m
          20
t₁ = 500sec

Now:

t₂  =S₂/ V₂

t₂ = 800
         4
t₂ = 200sec

Total time = t₁ + t₂
           = 500 + 200
           = 700sec

Now put values in eq. (1)
average velocity = 10,000 + 800
                                      70
                 = 10,800
                       70
average velocity = 15.4ms⁻¹

Given Data:
Initial velocity (vi) = 0ms⁻¹
Time (t) = 5sec
g = 10ms⁻²

To find:
height = s = ?
vf = ?

Solution:
For vf we use 1st equation of motion
vf = vi + gt
vf = 0 + (10)(5)
vf = 50ms⁻¹

For time we use 2nd equation of motion

S = Vit + 1/2 gt²

Putting values
S = (0)(5) + 1/2 (10)(5)²
S = 0 + 5(25)
S = 0 + 125
S = 125m

So, s = 125m , vf = 50ms⁻¹

Given data:
time (t) = 3sec
Vf = 0ms⁻¹
g = -10ms⁻²

To find:
initial velocity (vi) = ?
height = s = ?

Solution:
For initial velocity we use 1st equation of motion
Vf = Vi + at
Vf - gt = Vi
0 - (10)(3) = Vi
0 - (-30) = Vi
0 + 30 = Vi

Vi = 30ms⁻¹

For height we use 2nd equation of motion
S = Vit + (1/2)gt²
S = (30)(3) + (1)(-10)(3)²
                      2
S = 90 + (-5)(9)
S = 90 - 45
S = 45m

Since the ball was hit 1m so total height becomes
S = 45m + 1m
S = 46m