Class 11 Physics Chapter 3 Motion In A Straight Line
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NCERT Notes For Class 11 Physics Chapter 3 Motion In A Straight Line
Class 11 Physics Chapter 3 Motion In A Straight Line
CHAPTER THREE: MOTION IN A STRAIGHT LINE
MOTION
 Motion is change in position of an object with time.
 Branch of physics which deals with the motion of objects – Mechanics
 Mechanics is classified into
 i)Statics ii) Kinematics iii) Dynamics
 Statics deals with object at rest under the action of forces.
 In Kinematics, we study ways to describe motion without going into the causes of motion.
 Dynamics deals with objects in motion by considering the causes of motion.
POINT OBJECT
 If the size of the object is much smaller than the distance it moves, it is considered as point object.
Examples
 a railway carriage moving without jerks between two stations.
 a monkey sitting on top of a man cycling smoothly on a circular track.
FRAME OF REFERENCE
 A place from which motion is observed and measured is called frame of reference.
 Example: Cartesian coordinate system with a clock – the reference point at the origin.
TYPES OF MOTION
 Based on the number of coordinates required to describe motion, motion can be classified as:
 One dimensional motion (Rectilinear motion )
 Twodimensional motion
 Threedimensional motion.
One dimensional motion
 Motion along a straight line is called one dimensional motion or rectilinear motion.
 Only one coordinate is required to describe this motion.
 In onedimensional motion, there are only two directions (backward and forward, upward and downward) in which an object can move
Example:
 a car moving on a straight road.
 Freely falling body
Twodimensional motion
 Motion in a plane is called two dimensional motions.
 Two coordinates are required to represent this motion.
Example :
 A car moving on a plane ground
 A boat moving on a still lake
Threedimensional motion
Motion in a space is called three dimensional motion.
Three coordinates are required to represent this motion.
Example:
 Movement of gas molecules
 A flying bird
PATH LENGTH
 The length of the path covered by an object is called path length.
 It is the total distance travelled by the object.
 Path length is a scalar quantity — a quantity that has a magnitude only and no direction.
 For example, the path length of the car moving from O to P and then from P to Q is 360+120 = 480 m.
DISPLACEMENT
 It is the change of position in a definite direction.
 Displacement is a vector quantity –have both magnitude and direction.
 It can be positive, negative or zero.
 In one dimensional motion direction, the two directions can be represented using positive (+) and negative () signs.
 If x_{1} and x_{2} are the positions of an object in time t_{1} and t_{2}, the displacement in time interval
Δt = t_{2 }− t_{1} , is given by
Δx = x_{2 } − x_{1}
 If x2 > x1 , displacement is positive
 if x2 < x1 , displacement is negative.
 The magnitude of displacement may or may not be equal to the path length traversed by an object.
 If the motion of an object is along a straight line and in the same direction, the magnitude of displacement is equal to the total path length.
Uniform motion
 If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line.
AVERAGE VELOCITY
 Ratio of total displacement to the total time.
 The SI unit for velocity is m/s or m s–1
 The unit km h^{–1} is used in many everyday applications
 Average velocity is a vector quantity
 Average velocity can be positive or negative or zero.
 Slope of the DisplacementTime graph gives the average velocity.
AVERAGE SPEED
 Ratio of total path length travelled to the total time interval
 Average speed over a finite interval of time is greater or equal to the magnitude of the average velocity
 If the motion of an object is along a straight line and in the same direction, the magnitude of average velocity is equal to average speed.
 SI unit of average speed is same as that of velocity.
PROBLEM
 A car is moving along a straight line, It moves from O to P in 18 s and returns from P to Q in 6.0 s. What are the average velocity and average speed of the car in going (a) from O to P ? and (b) from O to P and back to Q ?
Solution
 Average velocity
 Average velocity
INSTANTANEOUS VELOCITY ( VELOCITY)
 It is the velocity of an object at an instant.
 It is the average velocity as the time interval tends to zero
 For uniform motion, velocity is the same as the average velocity at all instants.
INSTANTANEOUS SPEED (SPEED)
 Instantaneous speed or simply speed is the magnitude of velocity
 Instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant because for an infinitesimally small time interval, the motion of a particle can be approximated to be uniform.
AVERAGE ACCELERATION
 Ratio of change in velocity to time interval
 Where v_{2} and v_{1 }are the instantaneous velocities or simply velocities at time t_{2} and t_{1} .
 SI unit is m/s^{2}.
 Slope of the velocitytime graph gives average acceleration.
INSTANTANEOUS ACCELERATION (ACCELERATION)
 It is the acceleration at an instant.
 It is the average acceleration as the as the time interval tends to zero
 The instantaneous acceleration is the slope of the tangent to the v–t curve at that instant.
 Acceleration can be positive, negative or zero.
 It is a vector quantity.
GRAPHS RELATED TO MOTION
POSITIONTIME GRAPH ( x –t Graph)
 It is the graph drawn taking time along xaxis and position along yaxis
 Slope of the xt graph gives the average velocity.
 Slope of the tangent at a point in the xt graph gives the velocity at that point.
Uses of Position –Time Graph
 To find the position at any instant
 To find the velocity at any instant
 To obtain the nature of motion
Position time graph of stationary object
Position time graph of an object in uniform motion
Positiontime graph of a car
 The car starts from rest at time t = 0 s from the origin O and picks up speed till t = 10 s and thereafter moves with uniform speed till t = 18 s. Then the brakes are applied and the car stops att = 20 s and x = 296 m.
Positiontime graph of an object moving with positive velocity
Positiontime graph of an object moving with negative velocity
Positiontime graph for motion with positive acceleration
Positiontime graph for motion with negative acceleration
Positiontime graph for motion with zero acceleration
PROBLEM
• Calculate the average velocity between 5s and 7s from the graph.
Solution
PROBLEM2
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the xt graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Solution
Time taken to fall in pit =37s.
VELOCITY – TIME GRAPH ( vt GRAPH)
 A graph with velocity along Y –axis and time along Xaxis.
 The acceleration at an instant is the slope of the tangent to the v–t curve at that instant.
 Area under the vt graph gives the displacement.
Uses of vt graph
 To find the displacement
 To find the velocity at any time
 To find the acceleration at any time
 To know the nature of motion
vt graph of motion in positive direction with positive acceleration
vt graph of motion in positive direction with negative acceleration
vt graph of motion in negative direction with negative acceleration
vt graph of motion of an object with negative acceleration that changes direction at time t1.
PROBLEM1
Draw vt graph from the given xt graph.
Solution
PROBLEM2
 Velocitytime graph of a ball thrown vertically upwards with an initial velocity is shown in figure.
 What is the magnitude of initial velocity of the ball?
 Calculate the distance travelled by the ball during 20 s, from the graph.
 Calculate the acceleration of the ball from the graph
Solution
 100 m/s
3. Acceleration = slope of the graph
 Therefore acceleration = 10m/s^{2}
ACCELERATION –TIME GRAPH
 A graph with acceleration along Y –axis and time along Xaxis.
 Area under acceleration – time graph gives velocity.
PROBLEM
 The graph shows the velocity – time graph of a moving body in a one dimensional motion. Draw the corresponding acceleration – time graph
Solution
PROBLEM 2
Draw acceleration –time graph from the velocitytime graph given below.
Solution
PROBLEM 3
which of these cannot possibly represent onedimensional motion of a particle.
Solution
 No – because a particle cannot have two positions at the same instant of time.
 No – because particle can never have two values of velocities at the same instant of time.
 No speed cannot be negative
 No – total path length cannot decrease with time.
NATURE OF GRAPHS IN A NUTSHELL
xt Graph 
vt Graph 
at Graph 









KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION
velocity–time graph of an object moving with uniform acceleration and with initial velocity v_{0 }
Velocity – Time Relation
 We have
 Where v final velocity, a – acceleration v_{0} –initial velocity
at = v − v_{0 }
Or v = v_{0} + at
DisplacementTime Relation
 We know , area under vt graph = Displacement
 Thus, the displacement at any time interval 0 and t, is given y
 But v − v_{0} = at
 Thus
 Therefore
 If x_{o} is the initial displacement
Velocity –Displacement Relation
 We have
 Thus
Displacement = Average velocity x time
 Therefore
 But
 Thus
Therefore
v^{2} = v_{0}^{2 }+ 2ax
 If x_{o} is the initial displacement
v^{2} = v_{0}^{2} + 2a (x – x_{0} )
Thus the equation of motion are
PROBLEM
 A ball is thrown vertically upwards with a velocity of 20 m s–1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground.
a) How high will the ball rise ?
b) how long will it be before the ball hits the
ground? Take g = 10 m s–2
Solution
a) Given v_{0} = +20 m/s , a = g =10 m/s , v = 0
 Using the equation
v^{2} = v_{0}^{2} + 2a(y – y_{0})
 We get
(y − y_{0} ) = 20m
b). We have y_{0} = 25 m, y = 0 m, v_{o} = 20 m /s, a = –10m /s^{2},
 Using the equation
 Solving this quadratic equation we get, t=5s.
MOTION OF AN OBJECT UNDER FREE FALL
 A body falling under the influence of acceleration due to gravity alone is called free fall (air resistance neglected)
 If the height through which the object falls is small compared to the earth’s radius, g can be taken to be constant, equal to 9.8 m s–2.
 Free fall is an example of motion with uniform acceleration.
 Since the acceleration due to gravity is always downward, it is in the negative direction.
 Acceleration due to gravity = – g = – 9.8m/s2.
Equations of motion of a freely falling body
 For a freely falling body with v0=0 and y0 =0, the equations of motion are
Acceleration –Time graph of a freely falling body
Velocity – Time graph of a freely falling body
Position –Time graph of a freely falling body
Galileo’s law of odd numbers
 The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7……]
Proof
 Divide time interval of motion into equal intervals
 The distance travelled is found out using
 Thus ratio of distances is found to be 1:3:5:7:…..
STOPPING DISTANCE OF VEHICLES
 When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance.
 Stopping distance is an important factor considered in setting speed limits, for example, in school zones
 Stopping distance depends on the initial velocity (v0) and the braking capacity, or deceleration (–a) that is caused by the braking.
Equation for Stopping Distance
 Let the distance travelled by the vehicle before it stops be, d.
 Substituting v=0 , x=d and acceleration = – a in the equation
 Thus, stopping distance ,
 Thus stopping distance is proportional to square initial velocity.
REACTION TIME
 Reaction time is the time a person takes to observe, think and act.
 Dropping a ruler the reaction time can be calculated using the formula
 Where d is the distance moved before reaction.
RELATIVE VELOCITY
 It is the velocity measured whenever there is a relative motion between objects.
 The velocity of object B relative to object A is
v_{BA} = v_{B} – v_{A}
 Similarly, velocity of object A relative to object B is:
v_{AB} =v_{A} – v_{B}
 Thus, v_{BA }= v_{AB }
Special cases:
 If v_{B} = v_{A}, – relative velocity v_{AB} or v_{BA} is zero
 If v_{A} > v_{B}, v_{B} – v_{A }is negative, thus, object A overtakes object B at this time
Positiontime graphs of two objects with equal velocities
Positiontime graphs of two objects with unequal velocities
Positiontime graphs of two objects with velocities in opposite directions
PROBLEM
Two parallel rail tracks run north south. Train A moves north with a speed of 54 km h–1, and train B moves south with a speed of 90 km h–1. What is the
 Velocity of B with respect to A ?,
 b) Velocity of ground with respect to B ?
 c) velocity of a monkey running on the roof of the train A against its motion (with a velocity of 18 km h–1 with respect to the train A) as observed by a man standing on the ground ?
Solution