Derive the equation for position - time relation (S=ut+1/2 at² ) by graphs method
Answers
As per the second equation of motion, we know that,
S = ut + 1/2 at²
According to the graph
Distance = area under v-t graph
Distance = area of OABD
Distance=area of OABD rectangle+area of ∆ ABC
.°. S = OA × OD + 1/2 × AC × BC
S = u × t + 1/2 × t × (v–u)............ ➊
From first equation of motion
V = u + at
v – u = at
Now, putting v–u = at in equation ➊
S = ut + 1/2 × t × (at)
.°. S = ut + 1/2 at²
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Answer:
As per the second equation of motion, we know that,
S = ut + 1/2 at²
According to the graph
Distance = area under v-t graph
Distance = area of OABD
Distance=area of OABD rectangle+area of ∆ ABC
.°. S = OA × OD + 1/2 × AC × BC
S = u × t + 1/2 × t × (v–u)............ ➊
From first equation of motion
V = u + at
v – u = at
Now, putting v–u = at in equation ➊
S = ut + 1/2 × t × (at)
.°. S = ut + 1/2 at²
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