1.A cyclist driving at 5 m/s, picksup velocity of 10 m/s over a distance of 15 m.Calculate acceleration and time in which the cyclist picksup above velocity.
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GiveN :
- Initial velocity (u) = 5 m/s
- Final velocity (v) = 10 m/s
- Distance travelled (s) = 15 m
To FinD :
- Acceleration of cyclist
- Time interval in which it attains velocity of 10 m/s
SolutioN :
Use 3rd equation of motion :
v² - u² = 2as
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10² - 5² = 2 * a * 15
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100 - 25 = 30a
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75 = 30a
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a = 75/30
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a = 2.5
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Now, use 1st equation of motion :
v = u + at
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10 = 5 + 2.5 * t
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10 - 5 = 2.5t
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2.5t = 5
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t = 5/2.5
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t = 2 s
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Abhishek474241:
Perfect
Answered by
59
Explanation:
- Initial velocity(u) = 5m/s
- Final velocity(v) = 10m/s
- Distance covered(s) = 15m
- The acceleration and time in which the cyclist picks up the velocity.
According to the third equation of motion
Substituting the values:-
According to the first equation of motion:-
Substituting the values
Hence;
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