What is the ratio of distance Travelled by a body falling freely from rest in the first second and third seconds of its fall?

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Concept:

• Velocity: The rate of change of displacement of an object is velocity.
• Acceleration: The rate of change of velocity of an object is an acceleration.
• Equations of motion describe the behaviour of a physical system in terms of its motion as a function of time.

If an object is moving with constant acceleration 'a', initial velocity 'u', the distance travelled in tth second is given by:

\(s = u + {1\over 2}a (2t-1)\)

s is the distance travelled in tth second.

Calculation:

Given: Initial velocity of the body u = 0.

The motion under gravity so, the body is falling freely: a = g.

Distance covered in tth second:

\(s = u + {1\over 2}g (2t-1)\)

\(s = 0 + {1\over 2}g (2t-1)\)

⇒ \(s_{th} = {1\over 2}g (2t-1)\)

Distance travelled in the 1st second: t = 1;

\(s_{1} = {1\over 2}g (2\times1-1)\)

⇒ \(s_{1} = {g\over 2}\)

Distance travelled in the 2nd second: t = 2;

\(s_{2} = {1\over 2}g (2\times2-1)\)

⇒ \(s_{2} = {3\,g\over 2}\)

Distance travelled in the 3rd second: t = 3;

\(s_{3} = {1\over 2}g (2\times3-1)\)

⇒ \(s_{3} = {5\,g\over 2}\)

So, s1 : s2 : s3 = 1 : 3 : 5

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