establish s= ut + a from velocity time graph for a uniform accelerated motion
Anonymous:
s = u + at ?
s = ut + 1/2at²
should i tell you the answers to these?
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
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
No but thank you..!!
The question was wrong, so I was asking from the one who asked this question.
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Displacement of the particle in time (t) S = area under υ − t graph S = area OABC S = area of rectangle AODC + area of ∆ ADB
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Answered by
76
Answer:
Let ‘u’ be the initial velocity, ‘v’ be the final velocity and ‘t’ be the time taken by body to reach velocity ‘v’.
⟴ For second equation of motion:
Since area of v-t graph gives distance covered.
S = Area of triangle + Area of rectangle
➩S = 1/2 (v – u)t + ut
v – u = at (from first equation)
➩S = ut + 1/2at²
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⟴ For first equation of motion:
Since slope of v-t graph gives acceleration.
So tanØ = a
We know that,
a = (v – u)/t
➩at = v – u
➩v = u + at
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Consider the attachment.
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