A marble rolling on a smooth floor has an initial velocity of 0.4 m/s. If the floor offers a retardation of 0.02 m/s 2 ,calculate the time it will take to come to rest.
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Given :
- Initial velocity, u = 0.4 m/s
- Acceleration, a = - 0.02 m/s²
- Final velocity, v = 0 m/s
To find :
- Time taken, t
According to the question,
➞ v = u + at
Where,
- v = Final velocity
- u = Initial velocity
- a = Acceleration
- t = Time
➞ 0 = 0.4 + (-0.02) × t
➞ 0 - 0.4 = - 0.02t
➞ - 0.4 = - 0.02t
➞ 0.4 ÷ 0.02 = t
➞ 20 = t
So,the time taken is 20 seconds.
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More formulas :-
Newton's first equation of motion :
- v = u + at
Newton's second equation of motion :
- s = ut + ½ at²
Newton's third equation of motion :
- v² = u² + 2as
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