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 â€“ 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 â€“ 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 )
- Two-dimensional motion
- Three-dimensional 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 one-dimensional 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
Two-dimensional 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
Three-dimensional 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 â€” 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 â€“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
Î”t = t_{2 }âˆ’ t_{1} , is given by
Î”x = x_{2 } âˆ’ 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 sâ€“1
- The unit km h^{â€“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 Displacement-Time 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 velocity-time 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 vâ€“t curve at that instant.
- Acceleration can be positive, negative or zero.
- It is a vector quantity.
GRAPHS RELATED TO MOTION
POSITION-TIME GRAPH ( x â€“t Graph)
- It is the graph drawn taking time along x-axis and position along y-axis
- Slope of the x-t graph gives the average velocity.
- Slope of the tangent at a point in the x-t graph gives the velocity at that point.
Uses of Position â€“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
Position-time 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.
Position-time graph of an object moving with positive velocity
Position-time graph of an object moving with negative velocity
Position-time graph for motion with positive acceleration
Position-time graph for motion with negative acceleration
Position-time graph for motion with zero acceleration
PROBLEM
â€¢ Calculate the average velocity between 5s and 7s from the graph.
Solution
PROBLEM-2
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 x-t 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 â€“ TIME GRAPH ( v-t GRAPH)
- A graph with velocity along Y â€“axis and time along X-axis.
- The acceleration at an instant is the slope of the tangent to the vâ€“t curve at that instant.
- Area under the v-t graph gives the displacement.
Uses of v-t 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
v-t graph of motion in positive direction with positive acceleration
v-t graph of motion in positive direction with negative acceleration
v-t graph of motion in negative direction with negative acceleration
v-t graph of motion of an object with negative acceleration that changes direction at time t1.
PROBLEM-1
Draw v-t graph from the given x-t graph.
Solution
PROBLEM-2
- Velocity-time 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 â€“TIME GRAPH
- A graph with acceleration along Y â€“axis and time along X-axis.
- Area under acceleration â€“ time graph gives velocity.
PROBLEM
- The graph shows the velocity â€“ time graph of a moving body in a one dimensional motion. Draw the corresponding acceleration â€“ time graph
Solution
PROBLEM -2
Draw acceleration â€“time graph from the velocity-time graph given below.
Solution
PROBLEM -3
which of these cannot possibly represent one-dimensional motion of a particle.
Solution
- No â€“ because a particle cannot have two positions at the same instant of time.
- No â€“ because particle can never have two values of velocities at the same instant of time.
- No- speed cannot be negative
- No â€“ total path length cannot decrease with time.
NATURE OF GRAPHS IN A NUTSHELL
x-t Graph | v-t Graph | a-t Graph |
KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION
velocityâ€“time graph of an object moving with uniform acceleration and with initial velocity v_{0 }
Velocity â€“ Time Relation
- We have
- Where v- final velocity, a â€“ acceleration v_{0} â€“initial velocity
at = v âˆ’ v_{0 }
Or v = v_{0} + at
Displacement-Time Relation
- We know , area under v-t graph = Displacement
- Thus, the displacement at any time interval 0 and t, is given y
- But v âˆ’ v_{0} = at
- Thus
- Therefore
- If x_{o} is the initial displacement
Velocity â€“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 â€“ x_{0} )
Thus the equation of motion are
PROBLEM
- A ball is thrown vertically upwards with a velocity of 20 m sâ€“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 sâ€“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 â€“ y_{0})
- We get
(y âˆ’ y_{0} ) = 20m
b). We have y_{0} = 25 m, y = 0 m, v_{o} = 20 m /s, a = â€“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 earthâ€™s radius, g can be taken to be constant, equal to 9.8 m sâ€“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 = â€“ g = â€“ 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 â€“Time graph of a freely falling body
Velocity â€“ Time graph of a freely falling body
Position â€“Time graph of a freely falling body
Galileoâ€™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: 7â€¦â€¦]
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:â€¦..
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 (â€“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 = â€“ 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} â€“ v_{A}
- Similarly, velocity of object A relative to object B is:
v_{AB} =v_{A} â€“ v_{B}
- Thus, v_{BA }= -v_{AB }
Special cases:-
- If v_{B} = v_{A}, â€“ relative velocity v_{AB} or v_{BA} is zero
- If v_{A} > v_{B}, v_{B} â€“ v_{A }is negative, thus, object A overtakes object B at this time
Position-time graphs of two objects with equal velocities
Position-time graphs of two objects with unequal velocities
Position-time 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 hâ€“1, and train B moves south with a speed of 90 km hâ€“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 hâ€“1 with respect to the train A) as observed by a man standing on the ground ?
Solution