Graphs Class 11 Notes | EduRev

Physics Class 11

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Class 11 : Graphs Class 11 Notes | EduRev

The document Graphs Class 11 Notes | EduRev is a part of the Class 11 Course Physics Class 11.
All you need of Class 11 at this link: Class 11

3. Graphs  :

(i) Straight line :
A linear relation between y & x represents a straight line.
General equation of straight line
y = mx + c
m ≡ slope of line 
c ≡y intercept i.e. where the line cuts the y-axis.
Slope is defined as the tan of  angle made by the  straight line with positive x-axis in anticlockwise 
direction.   

Graphs Class 11 Notes | EduRev        
m < 0 ⇒  θ > 90°  
m > 0  ⇒ θ < 90°
0° ≤ θ < 180°

Ex.23 Draw the graph for the equation :   2y = 3x + 2  
Graphs Class 11 Notes | EduRev
Sol. 2y = 3x + 2  ⇒  y = Graphs Class 11 Notes | EduRev    
Graphs Class 11 Notes | EduRev   ⇒ θ < 90°
c = +1 > 0  
⇒  The line will pass through (0, 1) 

Ex.24 Draw the graph for the equation : 2y + 4x + 2 = 0
 Sol.
2y + 4x + 2 = 0 ⇒ y = – 2x – 1
m = – 2 < 0  i.e., θ > 90°    
c = – 1 i.e., 
line will pass through (0, –1) 

  •  (i) If c = 0 line will pass through origin.
    Graphs Class 11 Notes | EduRev

(ii) y = c will be a line parallel to x axis.
(iii) x = c will be a line perpendicular to y axis   

Graphs Class 11 Notes | EduRev 

(ii) Parabola
A general quadratic equation represents a parabola.
y = ax2 + bx + c a ≠ 0
if a > 0 ; It will be a opening  upwards parabola.
if a < 0 ; It will be a opening downwards parabola.
if c = 0 ; It will pass through origin.

e.g. y = 4 x2  + 3x      Graphs Class 11 Notes | EduRev

Average velocity & instantaneous velocity from Position vs time graph
Average velocity from tto t
= Graphs Class 11 Notes | EduRev
=   tan θ  = slope of the chord AB   
Graphs Class 11 Notes | EduRev  
Graphs Class 11 Notes | EduRev
when t2 approaches t1 point B approaches Point A and the chord AB becomes tangent to the curve. Therefore 
vinstantaneous = Slope of the tangent x – t curve

(iii) Reading of Graph
(A) Reading x v/s t graphs

Explanation

(1)   Graphs Class 11 Notes | EduRev  Body is at rest at x0.


(2)  Graphs Class 11 Notes | EduRev   Body starts from origin and is moving with speed tan θ away from origin.


(3) Graphs Class 11 Notes | EduRev    Body starts from rest from origin and moves away from origin with increasing speed velocity and positive acceleration.   

(4)  Graphs Class 11 Notes | EduRev   Body starts from rest from x = x0 and moves away from origin with increasing  velocity or +ve acceleration.

(5) Graphs Class 11 Notes | EduRev Body starts from x = x0 and  is moving toward the origin with constant velocity passes throw origin after same time and continues to move away from origin.

(6) Graphs Class 11 Notes | EduRev    Body starts from rest at x = x0 and then moves with increasing speed towards origin  

∴ acceleration is –ve

 

7)  Body starts moving away from origin with some initial speed. Speed of  body is  decreasing till t1 and it becomes 0 momentarily of t = t1 and at this instant. Its reverses its direction  and move towards the origin with increasing speed. 


(8) Graphs Class 11 Notes | EduRev    Body starts from origin moves away from origin in the –ve x-axis at t = t1 with decreasing speed and at t= t1 it comes at rest momentarily, Reverses its direction  moves towards the origin the increasing speed. Crosses the origin at t = t2.


(9) Graphs Class 11 Notes | EduRevBody starts from origin from rest and moves away from origin with increasing  speed.


(B) V-t Graphs 

(1)  Graphs Class 11 Notes | EduRev   Body is always at rest.

(2)  Graphs Class 11 Notes | EduRev   Body is moving with constant velocity v0

(3) Graphs Class 11 Notes | EduRev    Body is at rest initially then it starts moving with its velocity increasing at a  constant rate i.e. body is moving with constant acceleration.

(4) Graphs Class 11 Notes | EduRev    Body starts its motion with initial velocity v0 and continues to move with its velocity  increasing at a constant rate i.e.  acceleration of the body is constant.

 

(5) Graphs Class 11 Notes | EduRev   Body starts its motion with initial velocity v0. Then it continues to move with its  velocity decreasing at a constant rate i.e. acceleration of the body is negative  and constant.  At t = t0 the body comes to rest instantaneously and reverses its direction of motion and then continues to move with decreasing velocity or increasing speed.
For 0 < t < t0    motion of the body is deaccelerated (∴ speed is decreasing)
t > t0  motion of the body is accelerated (∴ speed is increasing)


(6) Graphs Class 11 Notes | EduRev    Body is at rest initially. Then it starts moving with increasing velocity. As time  increases its velocity is increasing more rapidly. i.e. the moving with increasing  acceleration.

(7)  Graphs Class 11 Notes | EduRev   Body starts its motion with initial velocity v0. Its  velocity is decreasing with time and at t = t0 . It becomes zero after body reverse its direction of motion and continues to move with decreasing velocity or increasing speed. Since velocity  of the body is decreasing for whole motion. Therefore, its acceleration is negative.For 0 < t < t0    motion of the  body is deaccelerated (speed is decreasing) t > t0  motion of the body is accelerated (Q speed is increasing)
(C) Reading of a - t Graphs

(1) Graphs Class 11 Notes | EduRev    acceleration of the body is zero that means the body is moving constant velocity. 


(2) Graphs Class 11 Notes | EduRev    Acceleration of the body is constant and positive.

(3) Graphs Class 11 Notes | EduRev    Acceleration of the body is constant and negative


 

(4)   Graphs Class 11 Notes | EduRev  Initially the acceleration of  the body is zero. Then its acceleration is increasing 
at a constant rate.

(5)  Graphs Class 11 Notes | EduRev   The body starts accelerating(initial acceleration zero) at t = 0. Its acceleration is negative for whole of its motion and is decreasing at a constant rate.

(6) Graphs Class 11 Notes | EduRev Initially acceleration of the body is zero. Its acceleration is positive for whole of  its motion. Its acceleration is increasing for whole of its motion.

(IV) Drawing of graphs on the basis of given information.

(a) If acceleration of the body is zero.
(i) If the velocity of the body is v0 and it starts from origin.
 Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev  
(ii) If at t = 0, x = x0 then 
Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev  

(iii) If at t = 0,  x = – x0  then

Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev

b) If a body has constant acceleration :
 For this section

(i) u0, x0 & a0 are positive constants. 

(ii) uº initial velocity 

(iii) vº  velocity at any time t.
(iv) xº Position at any time t.
xi º initial position

(i) if u = 0, a = a0
   
Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev This is wrong because it suggest the body don't have some initial velocity

Graphs Class 11 Notes | EduRev
 
v = a0t
(ii) If u = u0 ,  a = a0

Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev  
(iii)   if u = u0,  a = – a0
  Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev
Graphs Class 11 Notes | EduRev
 

iv) if u = – u0 ,  a = + a0
x = xi – u0t + Graphs Class 11 Notes | EduRev
Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev   
 Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev

(v) If u = u0,  a = – a0
 x = xi – u0t – Graphs Class 11 Notes | EduRev

Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev

Ex.25 Draw the 
 (a) position vs time graph
 (b) velocity vs time graph
 (c) acceleration vs time graph
 for the following cases 

(i) If a body is projected vertically upwards with initial velocity u. Take the projection point to be origin and upward direction as positive.
  Graphs Class 11 Notes | EduRev

Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev
(ii) If a body is dropped from a height h above the ground. Take dropping point to be origin and upward direction as +ve.
  Graphs Class 11 Notes | EduRev  

Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev  
(iii) If a body is projected vertically upwards from a tower of height h with initial velocity u. Take the projection point to be origin and upward direction as +ve.

Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev    
(iv) A car starting from rest accelerates uniformly at 2 ms–2 for 5 seconds and then moves with constant speed acquired for the next 5 seconds and then comes to rest retarding at 2 ms–2.
Draw its
(a) Position vs time graph (b) Velocity vs time graph (c) acceleration vs time graph

acceleration vs time graph
Graphs Class 11 Notes | EduRev
Graphs Class 11 Notes | EduRev


acceleration vs time graph   velocity vs time graph Position vs time graph

(v) A particle starts from x = 0 and initial speed 10 ms–1 and moves with constant speed 10ms–1 for 20 sec. and then retarding uniformly comes to rest in next 10 seconds.
acceleration vs time graph
Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev
Graphs Class 11 Notes | EduRev

(V) Conversion of velocity v/s time graph to speed v/s time graph.

As we know that magnitude of velocity represent speed therefore whenever velocity goes 

–ve take its mirror image about time axis.
 (VI) Conversion of displacement vs time graph to distance vs time graph   For distance time graph just make the mirror image of the displacement time graph from point of zero velocity onwards.

Graphs Class 11 Notes | EduRev

(VII) Conversion of v - t graphs in to x-t and a-t graphs
(i) Graphs Class 11 Notes | EduRev
(ii) Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev

(iii)  Graphs Class 11 Notes | EduRevGraphs Class 11 Notes | EduRev

(iv) x – t graph
From t = 0 to t = t1 acceleration = 0 therefore 
from t = 0 to t = t1, x - t graph will be a straight line.
From t = t1 to t2 acceleration is negative 

∴ It will be an opening downward parabola  Graphs Class 11 Notes | EduRev
Graphs Class 11 Notes | EduRev 

 

(v) upto t = t1 acceleration is +ve
t1 < t < tacceleration is zero.
t > t2 acceleration is –ve 
Graphs Class 11 Notes | EduRev
x - t graph
Graphs Class 11 Notes | EduRev
Some important points : 

  • Graphs Class 11 Notes | EduRev

 ⇒ Δv = area under the a - t curve 

  • Graphs Class 11 Notes | EduRev

⇒ Δx = area under the v - t curve

⇒ displacement = area under the v - t curve

Ex-26 If at t = 0 u = 5 ms–1 then velocity at t = 10 sec 

Graphs Class 11 Notes | EduRev
= u + change in velocity
= 5 + area of the shaded part
= 5 + 10 × 5
= 55 ms–1

Ex-27 if at t = 0, u = 2 ms–2 find out it maximum velocity

Graphs Class 11 Notes | EduRev
Since whole motion is accelerating. Therefore velocity  will be max at the end of the motion which will be
  Graphs Class 11 Notes | EduRev 

Ex-28 if at t = 0, u = 4 ms–1 
Find out v at
t = 10 sec,  t = 20 sec & t = 30 sec.
Since for whole motion acceleration of the body is positive  
Graphs Class 11 Notes | EduRev  

Graphs Class 11 Notes | EduRev

∵ x2 > x1
∴ At the end of the motion A is at a greater distance from the starting point then B

Ex-29 

Graphs Class 11 Notes | EduRev
 Conclusions :

(i) Both A & B starts their motion at same time t = 0 and from same point x = 0.
(ii) Both are moving away from the starting point.
(iii) A is moving with constant velocity while B starts its motion from rest and its velocity is increasing with time i.e. it has some positive acceleration.
(iv) Q At t = t1 the tangent on B's graph becomes parallel to the A's graphs 
∴ At t = t1 velocity of both A & B is same.
(v) For t < t1 velocity of A is greater than velocity of B. Therefore up to t = t1, separation between A & B increases with time.
(vi) For t > t1 velocity of B is greater than velocity of A. Therefore after t = t1 separation between A & B starts decreasing and it becomes zero at t = t2 where B overtakes A.

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