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Derivation of the three equation s of motion
Asked by mohini.ray | 05 Feb, 2020, 06:53: AM

(1) First equation of Motion: v = u + a t

Consider a body of mass m having initial velocity u.

Let after time t its final velocity becomes v due to uniform acceleration a.

Now we know that:
Acceleration = Change in velocity / Time taken

Acceleration = (Final velocity - Initial velocity) / Time taken

a = v - u / t
a t = v - u

or  v = u + a t

This is the first equation of motion.

(2) Second equation of motion: s = u t + 1/2 at2

Let the distance travelled by the body be s.

We know that
Distance = Average velocity x Time

Also, Average velocity = (u + v) / 2

.:  Distance (t) = (u + v) / 2 t            .eq.(1)

Again we know that:

v = u + at

Substituting this value of v in eq.(1), we get

s = (u + u + a t) / 2 t
s = (2 u + a t) / 2 t
s = (2 u t + a t2) / 2
s = (2 u t / 2) + (a t/ 2)

or  s = u t + (1/2) a t2

This is the 2nd equation of motion.

(3) Third equation of Motion:  v2 = u2 +2 a s

We know that

v = u + a t
v - u = a t

or  t = (v - u) / a                                    ..eq.(2)

Also we know that

Distance = average velocity x Time

.: s = [(v + u) / 2] x [(v - u) / a]

s = (v2  u2) / 2 a

2 a s = v2  u2

or  v2 = u2 + 2 a s

This is the third equation of motion.

Answered by Shiwani Sawant | 05 Feb, 2020, 10:08: AM

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