Motion can be of different types depending upon the type of path by which the object is going through:
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.
Sol. Given, π = 3.14, R = 10 mDistance = πR = 3.14 × 10 = 31.4 m
Displacement = 2 × R = 2 × 10 = 20 m
Example 2: A body travels 4 km towards North then he turns to his right and travels another 4 km before coming to rest. Calculate
(i) total distance travelled,
(ii) total displacement.
Sol. Total distance travelled = OA + AB = 4 km + 4 km = 8 km
Total displacement = OB
Conversion Factor
Change from km/hr to m/s = 1000m/(60×60)s = 5/18 m/s
Example: What will be the speed of body in m/s and km/hr if it travels 40 kms in 5 hrs?
Sol: 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 travels 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.
Sol: Speed during first half (v_{1}) = 40 km/hr
Speed during second half (v_{2}) = 20 km/hr
Average speed = (v_{1 }+ v_{2})/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.
Sol: 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
Example: A car speed increases from 40 km/hr to 60 km/hr in 5 sec. Calculate the acceleration of car.
Sol. 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 ms^{2}
Example: 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?
Sol. v = 0 km/hr, u = 20 km/hr, t = 0.5 hrs
Retardation, a = (vu)/t = (020)/0.5 = 200/5 = 40 km hr^{2}
Watch the animated video below to understand the concepts in easy manner.
(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 = (v_{2}  v_{1})/(t_{2}  t_{1})
But, v_{2}  v_{1}
∴ a = 0/(t_{2}  t_{1}) or a = 0
(ii) v/t graph for uniformly accelerated motion:
In uniformly accelerated motion, there will be an 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:
a_{1}' = a_{2}'
(v) v/t graph for nonuniformly decelerated motion:
Note: In v/t graph, the area enclosed between any two time intervals, t_{2}  t_{1}, will represent the total displacement by that body.
Total distance travelled by body between t_{2} and t_{1}
= 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 the s/t graph, which of the following body ‘A’ or ‘B’ will be faster?
Sol. v_{A} > v_{B}
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
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)
s = Area of trapezium OABC
Example 1: A car starting from rest moves with a uniform acceleration of 0.1 ms2 for 4 mins. Find the speed and distance travelled.
Sol: 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) = 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 car during this time.
Sol: Deceleration, a = − 6 ms^{2}; Time, t = 2 sec.
Distance, s = ?
Final velocity, v = 0 ms^{1} (∵ car comes to rest)
Now, v = u + at
⇒ u = v – at = 0 – (6×2) = 12 ms^{1}
s = ut + ½ at^{2 }= 12 × 2 + ½ (6 × 2^{2}) = 24 – 12 = 12 m
80 videos352 docs97 tests

1. What is the difference between distance and displacement? 
2. How does uniform and nonuniform motion differ? 
3. What is the difference between speed and velocity? 
4. What is acceleration? 
5. Can an object experience retardation or deceleration? 
80 videos352 docs97 tests


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