Physics

Example : 3
The distance between two points A and B is 100 m. A person moves from A to B with a speed of 20 m/s and from B to A with a speed of 25 m/s. Calculate average speed and average velocity.
Solution. (i) Distance from A to B = 100 m
Distance from B to A = 100 m
Thus, total distance = 200 m
Time taken to move from A to B, is given by
t₁ = = 100/20 = 5 seconds
Time taken from B to A, is given by
t₂ = = 100/25 = 4 seconds
Total time taken = t₁ + t₂ = 5 + 4 = 9 sec.
∴ Average speed of the person = = 200/9 m/s = 22.2 m/s
(ii) Since person comes back to initial position A, displacement will be zero, resulting zero average velocity.
Example : 4
A car moves with a speed of 40 km/hr for first hour, then with a speed of 60 km/hr for next half hour and finally with a speed of 30 km/hr for next 1 hours. Calculate the average speed of the car.
Solution. Distance travelled in first hour, is given by s₁ = speed × time = 40 km/hr × 1 hr = 40 km
Distance travelled in next half an hour, is given by
s₂ = speed × time = 60 km/hr × hr = 30 km
Distance travelled in last 1 hours, is given by
s₃ = speed × time = 30 km/hr × 1 hr = 45 km
Thus, total distance travelled = s₁ + s₂ + s₃ = 40 + 30 + 45 = 115 km
Total time taken = 1 + + 1 = 3 hours
∴ Average speed = = 38.33 km/hr

Example : 5
Figure shows time distance graph of an object. Calculate the following :
(i) Which part of the graph shows that the body is at rest ?
(ii) Average speed in first 10 s.
(iii) Speeds in different parts of motion.

Distance-time graph
Solution. (i) The part BC shows that the body is at rest.
(ii) In first 10 seconds, distance travelled = 100m
Thus, average speed = = 100/10 = 10 m/s
(iii) Speed of the object in part AB is given by slope = 100/6 = 50/3 m/s
Speed of object in part BC = 0 m/s
Speed of the object in part CD = = 60/2 = 30 m/s
Speed of object in part DE = = 40/2 = 20m/s
Example : 6
Time-velocity graph of a particle is shown in Figure. Calculate the distance travelled in first seconds.

Solution. Distance travelled in first 8s is given by area OABCG
= area of rectangle OAMG + area of triangle BMC
= 8 × 60 + × 4 × 40 = 480 + 80 = 560 m.