This article is formulated according to the 3rd chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.
Mechanics is the oldest and fundamental branch of physics and it is the study of the state of rest as well as the state of motion of an object under the action of force. The study of mechanics is broadly classified into (i) Statics and (ii) Dynamics
It deals with bodies at rest under the action of system of force.
It deals with motion of a body under the action of force. Dynamics is again divided into (a) Kinematics and (b) Kinetics.
It deals with the description of motion without reference to the cause of motion.
It deals with what moves and what causes motion. Some of the terms used in describing motion are given below.
A particle is ideally just a piece or quantity of matter, having no linear dimensions but only position and mass.
An event is a physical process that occurs at a point in space and at an instant of time.
A person or equipment which can locate, record, measure and interpret an event is called an observer.
Frame of reference
It is the reference in which an observer sits and makes the observations. In order to specify the position, we need to use a reference point and set of axes. The choice of set of axes in a frame of reference depends on the situation.
Motion is change in position of an object with time.
Motion of objects along a straight line. Ex: A car moving along a straight road, A freely falling body.
A body is said to be at rest when it does not change its position with time.
It is the actual distance covered by a body in time 𝑡. It is also called as distance travelled.
- Path length is a scalar quantity.
- SI unit of path length is “𝑚𝑒𝑡𝑟𝑒” (𝑚). Dimensions are 𝑀0𝐿𝑇0.
- Path length depends on the actual path.
- Path length is always positive.
It is the shortest distance between the initial point and final point.
- It is vector quantity.
- SI unit of displacement is “𝑚𝑒𝑡𝑟𝑒”(𝑚). Dimensions are 𝑀0𝐿𝑇0.
- Displacement may be positive, negative and zero.
- Magnitude of the displacement can never be greater than path length.
- When a body moves in straight line displacement is equal to path length.
- It is independent of the actual path travelled and it denoted by ∆𝑥.
Difference between path length and displacement
|It is the actual distance covered by a body in time t.||It is also called as distance travelled It is the shortest distance between the initial point and final point.|
|Path length is a scalar quantity.||It is vector quantity.|
|Path length is always positive.||Displacement may be positive, negative and zero.|
|Path length is always greater than or equal to displacement.||Displacement is always less than or equal to path length.|
Speed is defined as the rate of change of position of a particle.
𝑆𝑝𝑒𝑒𝑑 = 𝑝𝑎𝑡ℎ 𝑙𝑒𝑛𝑔𝑡ℎ/𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 = 𝑥/𝑡
- Speed is a scalar quantity.
- Its SI unit is 𝑚𝑒𝑡𝑟𝑒 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 (𝑚𝑠−1). Dimensions are 𝑀0𝐿𝑇−1.
- It is always positive.
- Speed gives no indication about the direction of motion of the particle.
The average speed of a particle in motion is defined as the ratio of the total path length to the total time taken.
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 = 𝑡𝑜𝑡𝑎𝑙 𝑝𝑎𝑡ℎ 𝑙𝑒𝑛𝑔𝑡ℎ/𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
Instantaneous speed (speed)
It is defined as the limit of average speed as the time interval is infinitesimally small.
Velocity is defined as the rate of change of displacement of a body.
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡/𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝒗 = 𝒙/𝒕
- Velocity is a vector quantity.
- SI unit is 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 (𝑚𝑠−). Dimensions are 𝑀0𝐿𝑇−1.
- Velocity may be positive, negative or zero.
The average velocity of a particle in motion is defined as the ratio of total displacement to the total time taken.
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡/𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑣̅= (𝑥2 − 𝑥1)/(𝑡2 − 𝑡1)
Velocity is defined as the limit of average velocity as the time interval ∆𝑡 becomes infinitesimally small. 𝒗 = 𝐥𝐢𝐦 ∆𝒕→𝟎 (∆𝒙/∆𝒕) = 𝒅𝒙/𝒅t
- Instantaneous velocity is also called velocity.
- In a position-time graph, the instantaneous velocity at a point is the slope to the tangent drawn to the curve at that point.
- Instantaneous speed or speed is the magnitude of velocity.
If equal changes of displacement take place in equal intervals of time is called uniform velocity.
Note: When a body moves with uniform velocity, neither the magnitude nor the direction of the velocity changes.
Difference between speed and velocity
|It is defined as the ratio of the path length to the time taken.||It is defined as the ratio of displacement to the time taken.|
|Speed is a scalar quantity.||Velocity is a vector quantity.|
|It is always positive.||Velocity may be positive, negative or zero.|
It is defined as the rate change of velocity of a particle.
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦/𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝒂 = 𝒗 − 𝒗𝟎𝒕
- Acceleration is a vector quantity.
- SI unit is 𝑚𝑒𝑡𝑟𝑒 𝑝𝑒𝑟 𝑠𝑞𝑢𝑎𝑟𝑒 𝑜𝑓 𝑠𝑒𝑐𝑜𝑛𝑑 (𝑚𝑠−2) and dimensions are 𝑀0𝐿𝑇−2.
- Since velocity is a quantity having both magnitude and direction, Acceleration may result from a change in magnitude or a change in direction or changes in both.
- Acceleration can be positive, negative, or zero.
- The negative acceleration is called retardation or deceleration.
It is defined as the total change in velocity divided by the total time taken.
𝑎̅ = (𝑣2 − 𝑣1)/(𝑡2 − 𝑡1) = ∆𝑣∆𝑡
It is defined as the limit of the average acceleration as the time interval ∆𝑡 becomes infinitesimally small.
𝒂 = 𝐥𝐢𝐦 ∆𝒕→𝟎 ∆𝒗/∆𝒕 = 𝒅𝒗/𝒅𝒕
If the velocity of a body changes by an equal amount in equal intervals of time, however small these time intervals may be, is called uniform acceleration.
Graphical representation of motion
A diagrammatical representation of the variation of one quantity with respect to another quantity is called a graph.
It is a graph obtained by plotting the instantaneous positions of a particle versus time. The slope of the position time graph gives the velocity of the particle.
A graph of velocity versus time is called a velocity-time graph.
- The area under the v-t graph with the time axis gives the value of displacement covered in a given time.
- The slope of the tangent drawn on a graph gives instantaneous acceleration.
Uses of velocity-time (v-t) graph / Significance of velocity-time (v-t) graph
- It is used to study the nature of motion.
- It is used to find the velocity of the particle at any instant in time.
- It is used to derive the equations of motion.
- It is used to find displacement and acceleration.
The kinematic equation for uniformly accelerated motion
For uniformly accelerated motion, we can derive some simple equations that relate displacement (𝑥), time taken (𝑡), initial velocity (𝑣0), final velocity (𝑣), and acceleration (𝑎). These equations are called Kinematic equations for uniformly accelerated motion. The Equations are, (i) 𝑣 = 𝑣0 + 𝑎𝑡 (ii) 𝑥 = 𝑣0𝑡 + 1/2 𝑎𝑡2 (iii) 𝑣2 = 𝑣02 + 2𝑎𝑥
Derivation of equation of motion by graphical method
(i) 𝒗 = 𝒗𝟎 + 𝒂𝒕
Consider a particle in motion with initial velocity 𝑣0 and constant acceleration 𝑎.
Let 𝑣 be the final velocity of the body at time 𝑡.
From graph, slope = 𝐵𝐶/𝐴𝐶 = (𝐵𝐷 – 𝐶𝐷)/𝐴𝐶
But, 𝐶𝐷 = 𝑂𝐴 and 𝐴𝐶 = 𝑂𝐷
Slope = (𝐵𝐷 – 𝑂A)/𝑂𝐷 = 𝑣 − 𝑣0𝑡
But, slope of v-t graph gives the acceleration.
𝑎 = (𝑣 − 𝑣0)/𝑡
𝑎𝑡 = 𝑣 − 𝑣0
𝑣 − 𝑣0 = 𝑎𝑡
𝒗 = 𝒗𝟎 + 𝒂𝒕
(ii) 𝒙 = 𝒗𝟎𝒕 + 𝟏/𝟐𝒂𝒕𝟐
Consider a particle in motion with initial velocity 𝑣0 and constant acceleration 𝑎.
Let 𝑣 be the final velocity of the body at time 𝑡. From graph, 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑣 − 𝑡 𝑔𝑟𝑎𝑝ℎ
𝑥 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 𝑂𝐴𝐵𝐷
𝑥 = 𝑎𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 𝑂𝐴𝐶𝐷
𝑥 = [ (1/2) × 𝐴𝐶 × 𝐵𝐶] + [𝑂𝐷 × 𝑂𝐴]
𝑥 = (1/2)𝑡(𝑣 − 𝑣0) + 𝑡𝑣0
But, 𝑣 − 𝑣0 = 𝑎𝑡
𝑥 = (1/2)𝑡(𝑎𝑡) + 𝑣0𝑡
𝑥 = (1/2)𝑎𝑡2 + 𝑣0𝑡
𝒙 = 𝒗𝟎𝒕 + (𝟏/𝟐)𝒂𝒕𝟐
(iii) 𝒗𝟐 = 𝒗𝟎𝟐 + 𝟐𝒂𝒙
Consider a particle in motion with initial velocity 𝑣0 and constant acceleration 𝑎. Let 𝑣 be the final velocity of the body at time 𝑡. From the graph, 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑣 − 𝑡 𝑔𝑟𝑎𝑝ℎ
𝑥 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 𝑂𝐴𝐵𝐷
𝑥 = (1/2) (𝑂𝐴 + 𝐵𝐷)𝐴𝐶
𝑥 = (1/2) (𝑣0 + 𝑣)𝑡
But, 𝑣 − 𝑣0 = 𝑎𝑡 and 𝑡 = (𝑣 − 𝑣0)/𝑎
𝑥 = (1/2)(𝑣0 + 𝑣)(𝑣 − 𝑣0/𝑎)
𝑥 = (1/2)(𝑣2 − 𝑣02)/𝑎
2𝑎𝑥 = 𝑣2 − 𝑣02
𝑣2 − 𝑣02 = 2𝑎𝑥
𝒗𝟐 = 𝒗𝟎𝟐 + 𝟐𝒂𝒙
Note: The set above equations were obtained by assuming that at 𝑡 = 0, the position of the Particle 𝑥 is 0 (zero). When at 𝑡 = 0, If the position of the particle is at 𝑥0(𝑛𝑜𝑛 𝑧𝑒𝑟𝑜), then the equations are, (i) 𝑣 = 𝑣0 + 𝑎𝑡 (ii) 𝑥 − 𝑥0 = 𝑣0𝑡 + (1/2)𝑎𝑡2 (iii) 𝑣2 = 𝑣02 + 2𝑎(𝑥 − 𝑥0).
An object released near the surface of the earth is accelerated downward under the influence of the force of gravity. If the air resistance is neglected, then the motion of the body is known as free fall.
Acceleration due to gravity
The acceleration produced in an object due to gravity is called acceleration due to gravity, denoted by 𝑔. Free fall is an example of motion along a straight line under constant acceleration.
- Acceleration due to gravity is always a downward vector directed toward the center of the earth.
- The magnitude of 𝑔 is approximately 9.8𝑚𝑠2 near the surface of the earth.
- Acceleration due to gravity is the same for all freely falling bodies irrespective of their size, shape, and mass.
- The distance traversed by a body falling freely from rest during equal intervals of time are in the ratio 1: 3: 5: 7: … … …. this is known as Galileo’s law of ODD numbers.
Equations of motion under gravity
The motion of a freely falling body is in the Y-direction. If we take vertically upward as a positive Y-axis, acceleration is along the negative Y-axis, therefore 𝑎 = −𝑔. Then, (i) 𝑣 = 𝑣0 − 𝑔𝑡 (ii) 𝑦 = 𝑣0𝑡 – (1/2)𝑔𝑡2 (iii) 𝑣2 = 𝑣02 − 2𝑔𝑦
For a freely falling body the initial velocity, 𝑣0 = 0. Then, (i) 𝑣 = −𝑔𝑡 (ii) 𝑦 = −(1/2)𝑔𝑡2 (iii) 𝑣2 = −2𝑔𝑦.
The 𝑎 − 𝑡 graph, 𝑣 − 𝑡 graph, and 𝑦 − 𝑡 graph to a body released from rest at 𝑦 = 0 are as shown.
Note: (i) Stopping distance: When breaks are applied to a moving vehicle, the distance traveled before stopping is called stopping distance.
𝒅𝒔 = −𝒗𝟎𝟐/𝟐𝒂
It is an important factor for road safety, and it depends on initial velocity and deceleration (−𝑎).
(ii) Reaction time: When a situation demands immediate action, it takes some time before we really respond this time is called reaction time.
The relative velocity of body 𝐴 with respect to body 𝐵 is defined as the time rate of change of displacement of 𝐴 with respect to 𝐵.
Explanation: Consider two bodies A and B moving with constant velocity 𝑣𝐴 and 𝑣𝐵 respectively, along the positive X-axis. Let 𝑥𝐴(𝑡) and 𝑥𝐵(𝑡) be the position of 𝐴 and 𝐵 at any given instant of time 𝑡, then 𝑥𝐴(𝑡) = 𝑥𝐴(0) + 𝑣𝐴𝑡, 𝑥𝐵(𝑡) = 𝑥𝐵(0) + 𝑣𝐵𝑡.
The separation between 𝐴 and 𝐵 at time 𝑡 is,
𝑥𝐵(𝑡) − 𝑥𝐴(𝑡) = 𝑥𝐵(0) − 𝑥𝐴(0) + (𝑣𝐵 − 𝑣𝐴)𝑡.
Here, 𝑥𝐵(0) − 𝑥𝐴(0) is the separation between 𝐴 and 𝐵 at 𝑡 = 0, and (𝑣𝐵 − 𝑣𝐴) is the time rate of change of relative velocity of 𝐵 with respect to 𝐴, denoted by 𝑣𝐵𝐴. Hence, 𝒗𝑩𝑨 = 𝒗𝑩 − 𝒗𝑨.
Similarly, the velocity of A with respect to B is 𝒗𝑨𝑩 = 𝒗𝑨 − 𝒗𝑩 and it can be shown that 𝑣𝐴𝐵 = −𝑣𝐵𝐴.
When two bodies move with the same velocity in the same direction, then 𝑣𝐴 = 𝑣𝐵 and 𝑣𝐴 − 𝑣𝐵 = 0 and 𝑣𝐴𝐵 = 𝑣𝐵𝐴 = 0, then two bodies appear at rest with respect to each other. In this case, the relative velocity is minimum.
When two bodies move in the same direction with different velocities, If 𝑣𝐴>𝑣𝐵 then 𝑣𝐵𝐴 = 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 and 𝑣𝐴𝐵 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒.
When two bodies move in different velocities or the same velocities in opposite directions. The magnitude of the relative velocity of either of them with respect to the other is equal to the sum of the magnitude of their velocities. 𝑣𝐵𝐴 = 𝑣𝐴𝐵 = 𝑣𝐴 + 𝑣𝐵.
In this case relative velocity is maximum.
1) A car is moving along a straight line. It moves O to P in 18 second covering 360m and returns from P to Q in 6second by covering 120m. Calculate average velocity and average speed of a car in going (a) from O to P (b) from O to P and back to Q.
𝑂𝑃 = 360𝑚
𝑄𝑃 = 120𝑚 𝑂
𝑄 = 𝑂𝑃 − 𝑄𝑃 = 360 − 120 = 240𝑚
(a) Avg. velocity = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡/𝑡𝑖𝑚𝑒
𝑣 = 𝑂𝑃/𝑡 = 360/18 = 20𝑚𝑠−1
Avg. speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑡𝑖𝑚𝑒
Avg. speed = 360/18 = 20𝑚𝑠−1
(b) Avg. velocity = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡/𝑡𝑖𝑚𝑒
𝑣 = 𝑂𝑄/𝑡 = 240/24 = 10𝑚𝑠−1
Avg. speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑡𝑖𝑚𝑒 = (360 + 120)/24
Avg. speed = 480/24 = 20𝑚𝑠−1
2) A car moving along a straight line takes 5 second to increase its velocity from 15𝑚𝑠−1 to 30𝑚𝑠−1. What is the acceleration of the car? Also calculate the distance travelled by the car in 5 second.
Given 𝑡 = 5𝑠, 𝑣0 = 15𝑚𝑠−1, 𝑣 = 30𝑚𝑠−1, 𝑎 = ?, 𝑥 = ?
𝑣 = 𝑣0 + 𝑎𝑡
30 = 15 + (𝑎 × 5)
5𝑎 = 30 − 15
𝑎 = 15/5 = 3𝑚𝑠−2
𝑥 = 𝑣0𝑡 + (1/2)𝑎𝑡2 𝑥
= (15 × 5) + ((1/2) × 3 × 52)
𝑥 = 75 + (75/2)
𝑥 = 75 + 37.5 = 112.5𝑚
3) A car moving along a straight road increases its velocity from 10𝑚𝑠−1 to 30𝑚𝑠−1 in 4 seconds. Calculate (a) the acceleration of the car and (b) the distance travelled by the car in 4 seconds.
Given 𝑣0 = 10𝑚𝑠−1, 𝑣 = 30𝑚𝑠−1, 𝑡 = 4𝑠, 𝑎 = ?, 𝑥 = ?
𝑣 = 𝑣0 + 𝑎𝑡
30 = 10 + (𝑎 × 4)
4𝑎 = 30 − 10
𝑎 = 20/4 = 5𝑚𝑠−2
𝑥 = 𝑣0𝑡 + (1/2)𝑎𝑡2
𝑥 = (10 × 4) + ((1/2) × 5 × 42)
𝑥 = 40 + (80/2)
𝑥 = 40 + 40 = 80𝑚
4) A truck moving along a straight highway with a speed of 20𝑚𝑠−1 is brought to rest in 10𝑠. What is the retardation of the truck? How far will the truck travel before it comes to rest?
Given 𝑣0 = 20𝑚𝑠−1, 𝑣 = 0𝑚𝑠−1, 𝑡 = 10𝑠, 𝑎 =?, 𝑥 = ?
𝑣 = 𝑣0 + 𝑎𝑡
0 = 20 + (𝑎 × 10)
10𝑎 = −20
𝑎 = −20/10 = −2𝑚𝑠−2
𝑥 = 𝑣0𝑡 + (1/2)𝑎𝑡2 = (20 × 10) + ((1/2) × (−2) × 102)
𝑥 = 200 − 100
𝑥 = 100𝑚
5) A player throws a ball upwards with an initial speed of 29.4𝑚𝑠−1 (a) What is the direction of acceleration during the upward motion of the ball? (b) What is the velocity and acceleration at the highest point of its path? (c) To what height does the ball rise and after how long does the ball returns to the player’s hand?
Given 𝑣0 = 29.4𝑚𝑠−1
(a) The ball is moving under the gravity, so the direction of acceleration is vertically downwards and towards the centre of the earth.
(b) At the highest point, 𝑣 = 0𝑚𝑠−1 and acceleration is equal to acceleration due to gravity, 𝑎 = 9.8𝑚𝑠−2
(c) 𝑣2 = 𝑣02 + 2𝑎𝑥
0 = (29.4)2 + (2 × (−9.8) × 𝑥)
𝑥 = (29.4)2/(2 × 9.8) = 44.1𝑚
𝑣 = 𝑣0 + 𝑎𝑡
0 = 29.4 + ((−9.8) × 𝑡)
𝑡 = 29.4/9.8 = 3𝑠
Total time = time of ascent + time of decent = 3s +3s = 6s
6) A car is moving along a straight highway with a speed of 126kmph is brought to stop with in 200m. What is the retardation of the car and how long does it take for the car to stop?
Given 𝑣0 = 126𝑘𝑚𝑝ℎ
𝑣0 = (126 × 1000)/3600𝑚𝑠−1 = 35𝑚𝑠−1
𝑥 = 200𝑚
𝑣 = 0𝑚𝑠−1
𝑎 = ? , 𝑡 = ?
𝑣2 = 𝑣02 + 2𝑎𝑥
0 = 352 + (2 × 𝑎 × 200)
400𝑎 = −35 × 35
𝑎 = − (35 × 35)/400 = −3.06𝑚𝑠−2
𝑣 = 𝑣0 + 𝑎𝑡
0 = 35 + (−3.06) × 𝑡
𝑡 = 35/3.06 = 11.42𝑠
7) Two trains A and B of length 300m each are moving on two parallel tracks with a uniform speed of 54kmph in the same direction, with the train A ahead of B. The driver of train B decides to overtake A and accelerates by 2𝑚𝑠−2. If after 25s, the guard of train B just brushes past the driver of A. What original distance between them?
Given 𝑙𝐴 = 300𝑚, 𝑙𝐵 = 300𝑚, 𝑣𝐴 = 54𝑘𝑚𝑝ℎ = 15𝑚𝑠−1, 𝑣𝐵 = 15𝑚𝑠−1, 𝑎𝐴 = 0𝑚𝑠−2, 𝑎𝐵 = 2𝑚𝑠−2, 𝑡 = 25𝑠, Original distance, 𝑙 = ?
Distance travelled by A in 25s, 𝑥𝐴 = 𝑣𝐴𝑡 + ((1/2)𝑎𝐴𝑡2)
𝑥𝐴 = (15 × 25) + 0
𝑥𝐴 = 375𝑚.
Distance travelled by B in 25s, 𝑥𝐵 = 𝑣𝐵𝑡 + ((1/2)𝑎𝐵𝑡2)
𝑥𝐵 = (15 × 25) + ((1/2) × 2 × 25 × 25)
𝑥𝐵 = 375 + 625 = 1000
Distance travelled by B in 25s, 𝑥𝐵 = 𝑙𝐵 + 𝑙 + 𝑥𝐴 + 𝑙𝐴
1000 = 300 + 𝑙 + 375 + 300
1000 = 975 + 𝑙
𝑙 = 1000 − 975 = 25𝑚
8) The displacement (in metre) of a particle moving along x-axis is given by 𝑥 = 2𝑡2+3. Calculate (i) Average velocity between 𝑡 = 3𝑠 and 𝑡 = 5𝑠. (ii) Instantaneous velocity at 𝑡 = 5 𝑠 and (iii) Instantaneous acceleration.
Given 𝑥 = 2𝑡2+3
(i)Avg. velocity = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡/𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 When 𝑡1 = 3𝑠,
𝑥1 = 2(3)2+3 = 21𝑚
When 𝑡2 = 5𝑠, 𝑥2 = 2(5)2+3 = 53𝑚
Average velocity = (𝑥2 − 𝑥1)/(𝑡2 − 𝑡1)
𝑣̅ = (53 – 21)/(5 – 3) = 32/2 = 16 𝑚𝑠−1
(ii) Instantaneous velocity, 𝑣 = 𝑑𝑥/𝑑𝑡
𝑥 = 2𝑡2+3
𝑣 = 𝑑𝑥/𝑑𝑡 = 2(2)𝑡 + 0 = 4𝑡
When 𝑡 = 5𝑠, 𝑣 = 𝑑𝑥/𝑑𝑡 = 4(5) = 20 𝑚𝑠−1
(iii)Instantaneous acceleration, 𝑎 = 𝑑𝑣/𝑑𝑡
𝑣 = 4𝑡
𝑎 = 𝑑𝑣/𝑑𝑡 = 4𝑚𝑠−2
9) Obtain equations of motion for constant acceleration using method of calculus.
(i) By definition, 𝑎 = 𝑑𝑣/𝑑𝑡
𝑑𝑣 = 𝑎𝑑𝑡
Integrating both sides,
𝑣 − 𝑣0 = 𝑎[𝑡 − 0]
𝑣 − 𝑣0 = 𝑎𝑡
𝒗 = 𝒗𝟎 + 𝒂𝒕
(ii) Now, we have 𝑣 = 𝑑𝑥/𝑑𝑡 𝑑𝑥 = 𝑣𝑑𝑡 Integrating both sides,
(𝒙 − 𝒙𝟎) = 𝒗𝟎𝒕 + (𝟏/𝟐)𝒂𝒕𝟐
(iii) We have, 𝑎 = 𝑑𝑣/𝑑𝑡 = (𝑑𝑣/𝑑𝑥)(𝑑𝑥/𝑑t)
𝑎 = (𝑑𝑣/𝑑𝑥)𝑣
𝑎𝑑𝑥 = 𝑣𝑑𝑣
Integrating on both sides,
𝑎(𝑥 − 𝑥0) = (1/2)(𝑣2 − 𝑣02)
2𝑎(𝑥 − 𝑥0) = (𝑣2 − 𝑣02)
𝒗𝟐 = 𝒗𝟎𝟐 + 𝟐𝒂(𝒙 − 𝒙𝟎)
Important questions for exams.
1) When will the magnitude of displacement equal to the path length?
2) Define average speed.
3) Define instantaneous speed.
4) Define instantaneous velocity.
5) Define average velocity.
6) Define acceleration.
7) What is retardation?
8) What is the acceleration of a body moving with uniform velocity?
9) What does the slope of position-time graph represent?
10) What does the slope of velocity-time graph represent?
11) Draw v-t graph for motion in uniform acceleration.
1) Distinguish between distance travelled and displacement of a particle.
2) Distinguish between speed and velocity.
3) Define uniform velocity and uniform acceleration.
4) What is position time graph? Draw 𝑥−𝑡 graph for an object at rest.
5) Draw position time graph for (a) a particle at rest,(b) a body moving with uniform velocity.
6) Draw the position time graph of a particle moving with a) Positive acceleration. b) Negative acceleration.
7) Draw v-t graph for body moving in uniform acceleration.
8) Define relative velocity. When will the relative velocity of two bodies be zero?
9) Define relative velocity with an example.
1) Write the significance of v-t graph.
2) Derive the equation 𝑣 = 𝑣0 + 𝑎𝑡 with usual notation by using v-t graph.
3) Define relative velocity. When does the relative velocity become maximum and minimum if two particles are moving along a straight line?
1) What is v-t graph? Derive the equation 𝑣2 = 𝑣02 + 2𝑎𝑥 with usual notation by using v-t graph.
2) What is v-t graph? Derive the equation 𝑥 = 𝑣0𝑡 + (1/2)𝑎𝑡2 with usual notation by using v-t graph.
1) The displacement of the particle moving along x-axis is given by 𝑥 = 3𝑡3 − 5𝑡2 + 1 where x is in metre and t is in second. Calculate (i) Instantaneous velocity at 𝑡 = 2 𝑠 (ii) Instantaneous acceleration at 𝑡 = 3 𝑠
2) A body is thrown up with velocity of 78.4𝑚𝑠−1. Find how high it will rise and how much time it will take to return to its point of projection.
3) A ball thrown vertically upwards and it reaches a height of 90𝑚. Find the velocity with which it was thrown, and the height reached by the ball 7 second after it was thrown?
4) A body travelling with an initial velocity 36𝑚𝑠−1 comes to rest after travelling 90𝑚. Assuming the retardation to be uniform, find its value. What time does it take to cover that distance?
5) A car is moving along a straight highway with a speed of 108𝑘𝑚ℎ𝑟−1 is brought to stop with a distance 200𝑚. What is the retardation of the car? And how long does it take for the car to stop?
6) A car travels a distance from A to B at a speed of 40𝑘𝑚ℎ𝑟−1 and returns to A at the speed of 30𝑘𝑚 ℎ𝑟−1. What is the average speed for the whole journey?