Derive the relation s= ut + 1/2 a(t)2 graphically
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Distance travelled = Area under the line
= ut + ½ (v-u)t
Acceleration (a) = (v-u)/t and so (v-u) = at
Therefore,
Distance travelled (s) = ut + ½ (v-u)t = ut + ½ (at)t = ut + ½ at²
Thus,proved.
= ut + ½ (v-u)t
Acceleration (a) = (v-u)/t and so (v-u) = at
Therefore,
Distance travelled (s) = ut + ½ (v-u)t = ut + ½ (at)t = ut + ½ at²
Thus,proved.
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