please answer q. 11 with figure please!!!!!!
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uniform velocity :- it means acceleration = 0
it means
velocity is constant .
hence,
when body starts to go and reach the wall , covered distance is S and displacement is D
S = vt = 10 × 6 = 60 m
D = 10 × 6 = 60m
now after striking body moves backward with same speed and reach initial point then displacement D' and and distance S'.
[ direction is changed( just opposite direction ) so, velocity sign changed e.g V = -10m/s
D' = -10 × 6 = - 60m
S' = | D' | = 60m
now ,
total distance = S + S' = 60 + 60 = 120m
total displacement = D+ D' = 60 - 60 = 0
it means
velocity is constant .
hence,
when body starts to go and reach the wall , covered distance is S and displacement is D
S = vt = 10 × 6 = 60 m
D = 10 × 6 = 60m
now after striking body moves backward with same speed and reach initial point then displacement D' and and distance S'.
[ direction is changed( just opposite direction ) so, velocity sign changed e.g V = -10m/s
D' = -10 × 6 = - 60m
S' = | D' | = 60m
now ,
total distance = S + S' = 60 + 60 = 120m
total displacement = D+ D' = 60 - 60 = 0
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john44:
thanks alot
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Answer:
uniform velocity :- it means acceleration =
it means
velocity is constant
.
hence,
when body starts to go and reach the wall, covered distance is S and displacement is D
S = vt = 10 x 6 = 60 m D = 10 x 6 = 60m
now after striking body moves backward with same speed and reach initial point then displacement D' and and distance S.
[ direction is changed( just opposite direction ) so, velocity sign changed e.g V = -10m/s
D' = -10 6 = -60m S' = I D'| = 60m
now,
total distance = S+ S' = 60 + 60 = 120m = total displacement = D+ D' = 60 - 60 = 0
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