derivative all three equations of motion
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Derivation of the Equations of Motion
First equation: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
Third equation: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
second equation: = 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
First equation: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
Third equation: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
second equation: = 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
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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
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
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