Difference between Distance and Displacement
Example 1: A body travels in a semicircular path of radius 10 m starting its motion from point ‘A’ to point ‘B’. Calculate the distance and displacement.
Total distance travelled by body, S = ?
Given, π = 3.14, R = 10 mS = πR
= 3.14 × 10 m
= 31.4 m
Total displacement of body, D = ?
Given, R = 10 m
D = 2×R = 2×10 m = 20 m
Example 2: A body travels 4 km towards North then he turn to his right and travels another 4 km before coming to rest. Calculate (i) total distance travelled, (ii) total displacement.
Total distance travelled = OA + AB
= 4 km + 4 km = 8 km
Total displacement = OB
Nonuniform Motion
Two types of nonuniformmotion
Example: What will be the speed of body in m/s and km/hr if it travels 40 kms in 5 hrs ?
Distance (s) = 40 km
Time (t) = 5 hrs.
Speed (in km/hr) = Total distance/Total time = 40/5 = 8 km/hr
40 km = 40 × 1000 m = 40,000 m
5 hrs = 5 × 60 × 60 sec.
Speed (in m/s) = (40 × 1000)/(5×60 ×60) = 80/36 = 2.22 m/s
Example 1: During first half of a journey by a body it travel with a speed of 40 km/hr and in the next half it travels with a speed of 20 km/hr. Calculate the average speed of the whole journey.
Speed during first half (v1) = 40 km/hr
Speed during second half (v 2 ) = 20 km/hr
Average speed = (v1+v2)/2 = (40+60)/2 = 60/2 = 30
Average speed by an object (body) = 30 km/hr.
Example 2: A car travels 20 km in first hour, 40 km in second hour and 30 km in third hour.
Calculate the average speed of the train.
Speed in Ist hour = 20 km/hr
Distance travelled during 1st hr = 1×20= 20 km
Speed in 2nd hour = 40 km/hr
Distance travelled during 2nd hr = 1×40= 40 km
Speed in 3rd hour = 30 km/hr
Distance travelled during 3rd hr = 1×30= 30 km
Average speed = Total distance travelled/Total time taken
= (20+40+30)/3 = 90/3 = 30 km/hr
Retardation/Deaceleration
Example 1: A car speed increases from 40 km/hr to 60 km/hr in 5 sec. Calculate the acceleration of car.
u = 40km/hr = (40×5)/18 = 100/9 = 11.11 m/s
v = 60 km/hr = (60×5)/18 = 150/9 = 16.66 m/s
t = 5 sec
a = (vu)/t = (16.66  11.11)/5
= 5.55/5 = 1.11 ms2
Example 2: A car travelling with a speed of 20 km/hr comes into rest in 0.5 hrs. What will be the value of its retardation?
v = 0 km/hr
u = 20 km/hr
t = 0.5 hrs
Retardation, a’ = (vu)/t = (020)/0.5
= 200/5 = 40 km hr2
VelocityTime Graph (v/t graph)
Note: The area enclosed between any two time intervals is ‘t_{2}  t_{1}’ in v/t graph will represent the total displacement by that body.
Total distance travelled by body between t_{2} and t_{1}, time intervals
= Area of ∆ABC + Area of rectangle ACDB
= ½ × (v_{2} – v_{1})×(t_{2}  t_{1}) + v_{1}× (t_{2}  t_{1})
Example: From the information given in s/t graph, which of the following body ‘A’ or ‘B’ will be more faster?
v_{A} > v_{B}
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 at^{2} (∵a = (vu)/t)
Third Equation: v^{2} = u^{2} + 2as
s = Area of trapezium OABC
⇒ v^{2} = u^{2} + 2as
Example 1: A car starting from rest moves with uniform acceleration of 0.1 ms2 for 4 mins. Find the speed and distance travelled.
u = 0 ms^{1} (∵ car is at rest)
a = 0.1 ms^{2}
t = 4 × 60 = 240 sec.
v = ?
From, v = u + at
v = 0 + (0.1 × 240)
⇒ v = 24 ms^{1 }
Example 2: The brakes applied to a car produces deceleration of 6 ms 2 in opposite direction to the motion. If car requires 2 sec. to stop after application of brakes, calculate distance travelled by the car during this time.
Deceleration, a = − 6 ms2
Time, t = 2 sec.
Distance, s = ?
Final velocity, v = 0 ms^{1} (∵ car comes to rest)
Now, v = u + at
Or u = v – at
Or u = 0 – (6×2) = 12 ms^{1}
And, s = ut + ½at^{2}
= 12 × 2 + ½ (6 × 2^{2})
= 24 – 12 = 12 m
80 videos352 docs97 tests

1. What is motion? 
2. What is the difference between distance and displacement? 
3. What is the difference between speed and velocity? 
4. What is the difference between uniform motion and nonuniform motion? 
5. How is acceleration related to motion? 
80 videos352 docs97 tests


Explore Courses for Class 9 exam
