Question

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.

Answer 1

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