how to derive the equations of motion
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The first equation of motion is: v=u+at
Acceleration = change in velocity/Time taken
=> Acceleration = Final velocity-Initial velocity / time taken
=> a = v-u /t
=>at = v-u
or v = u + at
The second equation of motion is s=ut+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 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
The third equation of motion is: 2as=u2-v2
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
Hope it helps..!!
cheers pls!
Acceleration = change in velocity/Time taken
=> Acceleration = Final velocity-Initial velocity / time taken
=> a = v-u /t
=>at = v-u
or v = u + at
The second equation of motion is s=ut+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 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
The third equation of motion is: 2as=u2-v2
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
Hope it helps..!!
cheers pls!
Dharmendra2120:
☺☺
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