Question

Q.3 Derive three equation of motion mathematically.


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Answer 1

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(1) First equation of Motion:

V = u + at

soln.

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

=>at = v-u

or v = u + at

This is the first equation of motion.

—————————————-

(2) Second equation of motion:

s = ut + 1/2 at^2

sol.

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 X t …….eq.(1)

Again we know that:

v = u + at

substituting this value of “v” in eq.(2), we get

s = (u+u+at)/2 x t

=>s = (2u+at)/2 X t

=>s = (2ut+at^2)/2

=>s = 2ut/2 + at^2/2

or s = ut +1/2 at^2

This is the 2nd equation of motion.

……………………………………………………………

(3) Third equation of Motion

v^2 = u^2 +2as

sol.

We know that

V = u + at

=> v-u = at

or t = (v-u)/a ………..eq.(3)

Also we know that

Distance = average velocity X Time

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

=> s = (v^2 – u^2)/2a

=>2as = v^2 – u^2

or v^2 = u^2 + 2as

This is the third equation of motion.

hope it helps you

Answer 2

Answer:

Derivation of the Equations of Motion

v = u + at

Let us begin with the first equation, v=u+at. This equation only talks about the acceleration, time, the initial and the final velocity. Let us assume a body that has a mass “m” and 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

Therefore,  Acceleration = (Final Velocity-Initial Velocity) / Time Taken

Hence, a = v-u /t or at = v-u

Therefore, we have: v = u + at

v² = u² + 2as

We have, v = u + at. Hence, we can write t = (v-u)/a

Also, we know that, Distance = average velocity × Time

Therefore, for constant acceleration we can write: Average velocity = (final velocity + initial velocty)/2 = (v+u)/2

Hence, Distance (s) = [(v+u)/2]  × [(v-u)/a]

or  s = (v² – u²)/2a

or 2as = v² – u²

or v² = u² + 2as

s = ut + ½at²

Let the distance be “s”. We know that

Distance = Average velocity × Time. Also, Average velocity = (u+v)/2

Therefore, Distance (s) = (u+v)/2 × t

Also, from v = u + at, we have:

s = (u+u+at)/2 × t = (2u+at)/2 × t

s = (2ut+at²)/2 = 2ut/2 + at²/2

or s = ut +½ at²

Explanation:

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