GRAPH OF PARENT FUNCTIONS
Saturday, June 13, 2015
Friday, June 12, 2015
SOLUTION OF I.E IRODOV : PROBLEM IN GENERAL PHYSICS (FIRST FIVE QUESTIONS)
SOLUTION OF IRODOV (FIRST FIVE PROBLEMS)
Irodov Problem 1.1
There are two ways of solving this problem, i) from the point of view of an observer on the raft ii) from the point of view of a person on the ground. I will provide both these methods since, I have got a lot of questions on this problem from the readers.
Since the raft is floating on the river, it moves at the same velocity as the river. Let the velocity of the river be s. Let the velocity of the boat in still water be v. This means that the boat's downstream velocity as seen from the ground will be v+s and it upstream velocity as seen from the ground will be v-s.
i) Solution From Reference Frame on the Raft
As seen by an observer on the raft, when the boat moves upstream its velocity will be v (v+s-s) and it will be -v (-v+s-s) when its moving downstream. Here, positive sign means that the boat is moving away from the raft and negative sign means that the boat is moving towards the raft.
Figure 1 shows the Time-Displacement diagram of the boat and the raft as seen from the point of view of the raft. The slopes of the both upstream and downstream paths of the boat relative to the raft are vand -v. Let t1 be the time takes by the raft to meet the boat after it turns around. As seen from the figure, it is obvious that
Thus, thus the total time between the two meetings of the raft and the boat is
Now the total distance traveled by the raft is l. This means that,
ii) Solution From Reference Frame on the Ground
The Time-Displacement diagram as seen from the ground of the raft and the boat are shown in Figure 2.
The raft moves at a constant velocity of s downstream (the slope of the line is s) and hence its path is represented by a straight blue line in Figure 2. The boat moves at a speed v+s when going upstream, represented by a line of slope v+s in Figure 2. The boat turns around and moves at a velocity -v+s when going downstream, represented by a line of negative slope of v-s.
From Figure 2 we can see that,
Now the total distance traveled by the raft is l. This means that,
Mean velocity is total distance by total time. The v-t graph is shown in the figure beside.
Suppose that time of constant acceleration was t. During the initial acceleration phase thus the car will travel
and its final speed will be 
. This can also be calculated using the v-t diagram as the area under the line AB (triangle ABE).
The car will take the same time t to come to complete rest and during this deceleration phase it will travel
. This can be calculated in the v-t diagram as the area of the triangleCFD.
During the uniform motion phase the car travels at a speedand travels for the remaining time of 
. Thus , during the uniform motion phase it travels a distance of 
. This can be calculated as the area of the rectangle EBCF.
The total distance traveled is thus given by,
The average velocity during the entire time is thus given by,
The above equation is a quadratic equation which has two possible solutions,
Clearly we choose the -ive sign since t cannot exceed the total time. The uniform interval is thus given by,
a) The average velocity is distance traveled by time = 200cm/20sec = 10cm/sec
b) The maximum velocity is the point where the rate of change of distance i.e. the slope of the curve is the maximum. In the mid portion of the curve, the point moves 100cm in 4 seconds. Thus, the slope is100/4 = 25cm/sec - this is the maximum speed achieved.
c) At any time the instataneous speed (ds/dt) is the slope of the curve while the average speed (s/t) is the tangent of the angle of the line joining the origin to the point.
Let us first consider a time t0 during the acceleration phase of the point. This is when t0 is in the interval (0,10). Throughout this region, the slope of the curve will always be greater than that of the line joining the point to the origin since the curve is convex. This means that throughout the acceleration phase, s/t will always be less thands/dt.
Now lets us consider the region where, the speed is constant. As shown in the picture even in this region the tangent of the line joining the point to the can never catch up the slope of the curve.
However, when we consider a time that is in the decelerating region of the curve, as the instantaneous speed of point decreases, at some time the average speed will catch up with the instantaneous speed. In other words the tangent of angle of line connecting the point to the origin will be same as the slope of the curve. This is shown in the figure beside.
Since exact nature (equation) of the curve of the decelerating part of the curve is not provided (it could be exponentially decaying) it is not possible to mathematically determine the exact value of time at which this happens. However, geometrically one can draw a tangent at every point on the curve and see if it passes through the origin. When this happens it will be the point at which instantaneous speed is same as the average speed.
Some readers have commented that this probably happens at 16s as suggested in the answer. Certainly it seems that way but the exact answer cannot be determined mathematically unless the nature iof the curve is specified.
Since the raft is floating on the river, it moves at the same velocity as the river. Let the velocity of the river be s. Let the velocity of the boat in still water be v. This means that the boat's downstream velocity as seen from the ground will be v+s and it upstream velocity as seen from the ground will be v-s.
i) Solution From Reference Frame on the Raft
As seen by an observer on the raft, when the boat moves upstream its velocity will be v (v+s-s) and it will be -v (-v+s-s) when its moving downstream. Here, positive sign means that the boat is moving away from the raft and negative sign means that the boat is moving towards the raft.
Figure 1 shows the Time-Displacement diagram of the boat and the raft as seen from the point of view of the raft. The slopes of the both upstream and downstream paths of the boat relative to the raft are vand -v. Let t1 be the time takes by the raft to meet the boat after it turns around. As seen from the figure, it is obvious that
Thus, thus the total time between the two meetings of the raft and the boat is
Now the total distance traveled by the raft is l. This means that,
ii) Solution From Reference Frame on the Ground
The Time-Displacement diagram as seen from the ground of the raft and the boat are shown in Figure 2.
The raft moves at a constant velocity of s downstream (the slope of the line is s) and hence its path is represented by a straight blue line in Figure 2. The boat moves at a speed v+s when going upstream, represented by a line of slope v+s in Figure 2. The boat turns around and moves at a velocity -v+s when going downstream, represented by a line of negative slope of v-s.
From Figure 2 we can see that,
Now the total distance traveled by the raft is l. This means that,
Irodov Problem 1.2
Mean velocity is total distance by total time. Let the total distance traveled be d. The time taken to travel half the distance (d/2) at speed vo is d/(2vo). Now another d/2 distance remains to be traveled. Now let the time taken to travel this remaining distance be t. Then(t/2)v1 + (t/2)v2 = d/2. This means that t = d/(v1 + v2).
The total time traveled thus is, d/(2vo) + d/(v1 + v2). The total distance is d. Thus, the mean velocity is,
The total time traveled thus is, d/(2vo) + d/(v1 + v2). The total distance is d. Thus, the mean velocity is,
Irodov Problem 1.3
Mean velocity is total distance by total time. The v-t graph is shown in the figure beside.
Suppose that time of constant acceleration was t. During the initial acceleration phase thus the car will travel
and its final speed will be 
. This can also be calculated using the v-t diagram as the area under the line AB (triangle ABE).The car will take the same time t to come to complete rest and during this deceleration phase it will travel
. This can be calculated in the v-t diagram as the area of the triangleCFD.During the uniform motion phase the car travels at a speed
and travels for the remaining time of 
. Thus , during the uniform motion phase it travels a distance of 
. This can be calculated as the area of the rectangle EBCF.The total distance traveled is thus given by,
The average velocity during the entire time is thus given by,
The above equation is a quadratic equation which has two possible solutions,
Clearly we choose the -ive sign since t cannot exceed the total time. The uniform interval is thus given by,
Irodov Problem 1.4
a) The average velocity is distance traveled by time = 200cm/20sec = 10cm/sec
b) The maximum velocity is the point where the rate of change of distance i.e. the slope of the curve is the maximum. In the mid portion of the curve, the point moves 100cm in 4 seconds. Thus, the slope is100/4 = 25cm/sec - this is the maximum speed achieved.
c) At any time the instataneous speed (ds/dt) is the slope of the curve while the average speed (s/t) is the tangent of the angle of the line joining the origin to the point.
Let us first consider a time t0 during the acceleration phase of the point. This is when t0 is in the interval (0,10). Throughout this region, the slope of the curve will always be greater than that of the line joining the point to the origin since the curve is convex. This means that throughout the acceleration phase, s/t will always be less thands/dt.Now lets us consider the region where, the speed is constant. As shown in the picture even in this region the tangent of the line joining the point to the can never catch up the slope of the curve.
However, when we consider a time that is in the decelerating region of the curve, as the instantaneous speed of point decreases, at some time the average speed will catch up with the instantaneous speed. In other words the tangent of angle of line connecting the point to the origin will be same as the slope of the curve. This is shown in the figure beside.
Since exact nature (equation) of the curve of the decelerating part of the curve is not provided (it could be exponentially decaying) it is not possible to mathematically determine the exact value of time at which this happens. However, geometrically one can draw a tangent at every point on the curve and see if it passes through the origin. When this happens it will be the point at which instantaneous speed is same as the average speed.
Some readers have commented that this probably happens at 16s as suggested in the answer. Certainly it seems that way but the exact answer cannot be determined mathematically unless the nature iof the curve is specified.
Irodov Problem 1.5
The displacement vector of the first particle as a function of time is s1 = r1 + t.v1 and that of the second will be s2 = r2+ t.v2. When and if the particles collide, s1 must be equal to s2. Thus,
r1 + t.v1 = r2 + t.v2. Thus, r1-r2 = t.(v2-v1). Since t is only a scalar (1 dimensional) while r and v are vector (more than 1 dimensional), for this condition to be true, r1-r2 must be aligned in the same direction as v2-v1. Thus, (r1-r2)/|r1-r2| = (v2-v1)/|v2-v1|.
r1 + t.v1 = r2 + t.v2. Thus, r1-r2 = t.(v2-v1). Since t is only a scalar (1 dimensional) while r and v are vector (more than 1 dimensional), for this condition to be true, r1-r2 must be aligned in the same direction as v2-v1. Thus, (r1-r2)/|r1-r2| = (v2-v1)/|v2-v1|.
Thursday, June 11, 2015
tips to crack IIT-JEE
So, you wish to be in one of the IITs or NITs? Are you targeting to crack JEE Main 2015? Have you started your preparation?- Objective type question with multiple choices
- Each question carries 4 marks & one negative mark is deducted for each wrong attempt
- Paper I shall be comprised of Physics, Chemistry & Mathematics questions with equal weight-age to all 3 subjects.
- Time duration: 3 hours
Paper pattern for Paper II
- Paper II is being conducted for admissions into Architecture/Planning Courses.
- The entire paper is divided into three parts: Part I – Mathematics, Part II – Aptitude Test, Part III – Drawing Test
JEE Main 2015 Syllabus
Click here for complete syllabus of JEE Main 2015
JEE Main 2015 Paper Analysis
Important Subject wise Topics along with expected number of marks from each topic:
- Mathematics:
Name of topic
|
Marks
|
Complex Numbers and Quadratic Equations
|
12
|
Binomial Theorem and its Simple Applications
|
4
|
Sequences and Series
|
8
|
Co-ordinate Geometry (Straight Lines & Circles, Conic Sections)
|
20
|
Statistics and Probability
|
8
|
Trigonometry
|
8
|
Mathematical Reasoning
|
4
|
Sets, Relations and Functions
|
4
|
Matrices and Determinants
|
8
|
Limit, Continuity and Differentiability
|
4
|
Integral Calculus
|
12
|
Differential Equations
|
4
|
Three Dimensional Geometry & Vector Algebra
|
12
|
Differentiation and applications
|
12
|
- Physics:
Name of Topic
|
Marks
|
Units and Measurement
|
4
|
Kinematics
|
8
|
Laws of Motion
|
4
|
Work, Power and energy
|
4
|
Rotational Motion
|
8
|
Gravitation
|
4
|
Properties of Solids and Liquids
|
12
|
Thermodynamics
|
12
|
Oscillations and Waves
|
8
|
Electrostatics
|
8
|
Current Electricity
|
4
|
Magnetic Effects of Current and Magnetism
|
8
|
Electromagnetic Induction and Alternating Currents
|
4
|
Electromagnetic Waves
|
8
|
Optics
|
12
|
Atoms and Nuclei
|
8
|
Electronic Devices
|
4
|
- Chemistry:
Name of topic
|
Marks
|
Some Basic Concepts in Chemistry
|
12
|
States of Matter (Gaseous State, Liquid State)
|
4
|
Atomic Structure
|
4
|
Chemical Bonding and Molecular Structure
|
8
|
Chemical Thermodynamics
|
4
|
Equilibrium
|
4
|
Redox Reactions
|
4
|
Classification of Elements and Periodicity in Properties
|
4
|
Hydrogen
|
4
|
S-Block Elements (Alkali and Alkaline Earth Metals)
|
4
|
Some Basic Principles of Organic Chemistry
|
4
|
Hydrocarbons
|
4
|
Chemical Kinetics
|
4
|
Solid State
|
4
|
Electrochemistry
|
12
|
General Principles and Processes of Isolation of Metals
|
8
|
P-Block Elements
|
4
|
Co-ordination Compounds
|
8
|
Organic Compounds containing Halogens
|
8
|
Organic Compounds containing Oxygen
|
12
|
Important Books To be Referred
Please click here for the complete list of books for JEE Main preparation
Important Tips for JEE Main preparation
- Time Management: Maintaining a proper schedule for every task that you do can lead you to success. Prepare along with all daily tasks such as exercise, play and every other tasks by managing the time.
- Diet & food: Maintain a completely healthy diet including fresh fruits and vegetables. Try to ignore oily stuff and have a proper balanced diet.
- Positivity: Always try to be positive towards every work that you do and be focused towards all your activities along with studies. Being optimistic works very well many a times in such situations.
- Mock Tests: Prepare the best way by appearing as many mock tests you can, and analyze the results regularly. Try solving the portion which you find difficult in the mock tests and create a strong base.
- Smart way to Prepare: Make separate sheets for formulas and theorem that can help you a lot while solving numerical in Physics and Mathematics.
- Believe in yourself: Self-belief is most important as it generates a lot of positivity and gives boost to your confidence.
- Revise: Make sure you revise every topic thoroughly and prepare well every topic by revising time and again. Also prepare separate revision plan for last minute preparation
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