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}
(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
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