A cyclist is cycling at a uniform velocity of 8 m/s for 8 seconds. He then stops paddling and the cycle comes to rest in next ten seconds. Draw the v-t graph and calculate:
(i) the average retardation. (ii) the distance covered with uniform
velocity.
(iii) the distance covered with variable velocity. (iv) the average velocity of the cyclist.
Answers
Answer:
Initial velocity = u = 8 m/s
Time taken = 8 s
Final velocity = v = 0 m/s
Tike taken to come to stop = 10 s
The v-t graph is in the attached photo.
(I) Average Retardation
⇒ Change in velocity/Time taken to make the change
⇒ CD/DE
⇒ 8-0/10
⇒ 8/10
⇒ 0.8 m/s²
(ii) Distance covered with uniform velocity
⇒ Speed × Time
⇒ AB × BD
⇒ 8 m/s × 8 s
⇒ 64 m
(iii) Distance covered with variable velocity
⇒ Area of ∆ CDE
⇒ ½*CD*DE
⇒ ½*8*10
⇒ 4*10
⇒ 40 m
(iv) Average velocity of the cyclist
⇒ Total Displacement/Total Time
⇒ (64+40) m/(8+10) s
⇒ 104/18 s
⇒ 5.77 m/s
Hence,
- the average retardation = 0.8 m/s²
- the distance covered with uniform velocity = 64 m
- the distance covered with variable velocity = 40 m
- the average velocity of the cyclist = 5.77 m/s
Given :-
A cyclist is cycling at a uniform velocity of 8 m/s for 8 seconds. He then stops paddling and the cycle comes to rest in next ten seconds
To Find :-
(i) the average retardation. (ii) the distance covered with uniform
velocity.
(iii) the distance covered with variable velocity. (iv) the average velocity of the cyclist.
Solution :-
(i) Average retardation = v - u/t
= 0 - 8/8
= -8/8
= -1 m/s²
(ii) Distance = Speed × Time
Distance = 8 × 8
Distance = 64 m
(iii) Distance covered = Area of triangle
Area of triangle = 1/2 × b × h
Area = 1/2 × 8 × 10
Area = 1/2 × 80
Area = 40 m²
(iv) Average velocity = Displacement/Time
Avg. Velocity = 64 + 40/8 + 10
Avg. velocity = 104/18
Avg. velocity = 5.76 m/s
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