Physics, asked by saugatpandey7520, 1 year ago

Solve three eq. Of motion

Answers

Answered by komalsoliwal2
1

s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time of motion. These equations are referred as SUVAT equations where SUVAT stands for displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (T)


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


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