Graphs | Physics for EmSAT Achieve PDF Download

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.   

        
m < 0 ⇒  θ > 90°  
m > 0  ⇒ θ < 90°
0° ≤ θ < 180°

Ex.23 Draw the graph for the equation :   2y = 3x + 2  

Sol. 2y = 3x + 2  ⇒  y =     
   ⇒ 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.
  

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

 

(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      

Average velocity & instantaneous velocity from Position vs time graph
Average velocity from t₁ to t₂ 
=
=   tan θ  = slope of the chord AB   
  

when t₂ approaches t₁ point B approaches Point A and the chord AB becomes tangent to the curve. Therefore 
v_{instantaneous} = Slope of the tangent x – t curve

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

Explanation

(1)     Body is at rest at x₀.

(2)     Body starts from origin and is moving with speed tan θ away from origin.

(3)     Body starts from rest from origin and moves away from origin with increasing speed velocity and positive acceleration.   

(4)     Body starts from rest from x = x₀ and moves away from origin with increasing  velocity or +ve acceleration.

(5)  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)     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)     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) Body starts from origin from rest and moves away from origin with increasing  speed.

(B) V-t Graphs 

(1)     Body is always at rest.

(2)     Body is moving with constant velocity v₀

(3)     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)     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)    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)     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)     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)     acceleration of the body is zero that means the body is moving constant velocity. 

(2)     Acceleration of the body is constant and positive.

(3)     Acceleration of the body is constant and negative

(4)     Initially the acceleration of  the body is zero. Then its acceleration is increasing 
at a constant rate.

(5)     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)  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.
   
(ii) If at t = 0, x = x0 then 
  

(iii) If at t = 0,  x = – x0  then

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
   
 This is wrong because it suggest the body don't have some initial velocity

 
v = a₀t
(ii) If u = u₀ ,  a = a₀

  
(iii)   if u = u₀,  a = – a₀
 

 

iv) if u = – u₀ ,  a = + a₀
x = xi – u₀t + 
   
 

(v) If u = u₀,  a = – a₀
 x = x_i – u₀t – 

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.
  

(ii) If a body is dropped from a height h above the ground. Take dropping point to be origin and upward direction as +ve.
    

  
(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.

    
(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

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

(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.

(VII) Conversion of v - t graphs in to x-t and a-t graphs
(i)
(ii) 

(iii) 

(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 
 

(v) upto t = t₁ acceleration is +ve
t1 < t < t₂ acceleration is zero.
t > t₂ acceleration is –ve 

x - t graph

Some important points : 

 ⇒ Δv = area under the a - t curve 

⇒ 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 

= 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

Since whole motion is accelerating. Therefore velocity  will be max at the end of the motion which will be
   

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  
  

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

Ex-29 

 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.