Motion can be of different types depending upon the type of path by which the object is going through.
(i) Circulatory motion/Circular motion – In a circular path.
(ii) Linear motion – In a straight line path.
(iii) Oscillatory/Vibratory motion – To and fro path with respect to origin.
Difference between Distance and Displacement
Uniform Motion
When a body travels equal distance in equal interval of time, then the motion is said to be uniform motion.
Nonuniform Motion
In this type of motion, the body will travel unequal distances in equal intervals of time.
Two types of nonuniformmotion
(i) Accelerated Motion: When motion of a body increases with time.
(ii) Deaccelerated Motion: When motion of a body decreases with time.
• Change from km/hr to m/s = 1000m/(60×60)s = 5/18 m/s
It is the speed of a body in given direction.
For nonuniform motion in a given line, average velocity will be alculated in the same way as done in average speed.
It can be positive (+ve), negative (ve) or zero.
→ Deaceleration is seen in nonuniform motion during decrease in velocity with time. It has same definition as acceleration.
• Deaceleration (a') = Change in velocity/Time = (vu)/t
Here, v < u, ‘a’ = negative (ve).
DistanceTime Graph (s/t graph)
(i) s/t graph for uniform motion:
(ii) s/t graph for nonuniform motion:
(iii) s/t graph for a body at rest:
v = (s_{2}  s_{1})/(t_{2}  t_{1})
But, s_{2}  s_{1}
∴ v = 0/(t_{2}  t_{1}) or v=0
(i) v/t graph for uniform motion:
a = (v2  v1)/(t2  t1)
But, v2  v1
∴ a = 0/(t2  t1) or a = 0
(ii) v/t graph for uniformly accelerated motion:
In uniformly accelerated motion, there will be equal increase in velocity in equal interval of time throughout the motion of body.
(iii) v/t graph for nonuniformly accelerated motion:
a_{2} ≠ a_{1}
(iv) v/t graph for uniformly decelerated motion:
or, a_{1}' = a_{2}'
(v) v/t graph for nonuniformly decelerated motion:
a_{1}' ≠ a_{2}'
Note: The area enclosed between any two time intervals is ‘t2  t1’ in v/t graph will represent the total displacement by that body.
Total distance travelled by body between t2 and t1, time intervals
= Area of ∆ABC + Area of rectangle ACDB
= ½ × (v_{2} – v_{1})×(t_{2}  t_{1}) + v_{1}× (t_{2}  t_{1})
First Equation: v = u + at
Final velocity = Initial velocity + Acceleration × Time
Graphical Derivation
Suppose a body has initial velocity ‘u’ (i.e., velocity at time t = 0 sec.) at point ‘A’ and this velocity changes to ‘v’ at point ‘B’ in ‘t’ secs. i.e., final velocity will be ‘v’.
For such a body there will be an acceleration.
a = Change in velocity/Change in Time
⇒ a = (OB  OA)/(OC0) = (vu)/(t0)
⇒ a = (vu)/t
⇒ v = u + at
Second Equation: s = ut + ½ at^{2}
Distance travelled by object = Area of OABC (trapezium)
= Area of OADC (rectangle) + Area of ∆ABD
= OA × AD + ½ × AD × BD
= u × t + ½ × t × (v – u)
= ut + ½ × t × at
⇒ s = ut + ½ at2 (∵a = (vu)/t)
Third Equation: v2 = u2 + 2as
s = Area of trapezium OABC
80 videos363 docs97 tests

NCERT Exemplar: Motion Doc  8 pages 
Short & Long Answer Questions Motion Doc  4 pages 
Motion: Introduction & Graphs Video  22:28 min 
1. What is motion? 
2. How is motion measured? 
3. What are the different types of motion? 
4. What is the difference between speed and velocity? 
5. What is the relationship between distance, time, and speed in motion? 
NCERT Exemplar: Motion Doc  8 pages 
Short & Long Answer Questions Motion Doc  4 pages 
Motion: Introduction & Graphs Video  22:28 min 

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