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December 14, 2023 742 0
Objects can be at rest or in movement in everyday life. Motion is often inferred through indirect evidence, such as the movement of dust indicating air motion. The phenomena of sunrise, sunset, and changing of seasons are related to the motion of the Earth, even if we don’t directly perceive it.
Example: An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
Solution:
Total distance travelled by the object = 16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
Average speed = Total distance travelled/Total time taken
= 32 m /6s =5.33 m/s
Therefore, the average speed of the object is 5.33 m/s.
Example: Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Solution:
The total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in 1 min = 0 m
Average speed = Total distance covered / Total time taken
180m/1min = 180m/ 1min ×1min/60s = 3 m/s
Average velocity = Displacement / Total time taken
= 0m/60 s
= 0 m/s
The average speed of Usha is 3 m/s and her average velocity is 0 m/s.
Example. Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in 30 s. Then he applies brakes such that the veMotion and Measurement locity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases.
Solution:
In the first case:
initial velocity, u = 0 ; final velocity, v = 6 m s–1 ; time, t = 30 s .
a= v-u/ta
Substituting the given values of u, v and t in the above equation, we get
a= (6ms–1 – 0 ms–1) / 30 s
= 0.2 m s–2
In the second case:
initial velocity, u = 6 m s–1; final velocity, v = 4 m s–1; time, t = 5 s.
Then, a = (4ms–1 –6ms–1) / 5s
= – 0.4 m s–2 .
The acceleration of the bicycle in the first case is 0.2 m s–2 and in the second case, it is – 0.4 m s–2.
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