please......... answer
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is it 50m...............
displacement is the shortest distance which is calculated by taking the velocity with which the body has completed the shortest path......
displacement = velocity×time
displacement = 5×10= 50m
velocity of 5m/s is considered because the shortest distance between final and initial point is AE.
Velocity graph for AE represents a velocity of 5 m/s in total time 10s.....
so displacement = 5×10 =50m
...
displacement is the shortest distance which is calculated by taking the velocity with which the body has completed the shortest path......
displacement = velocity×time
displacement = 5×10= 50m
velocity of 5m/s is considered because the shortest distance between final and initial point is AE.
Velocity graph for AE represents a velocity of 5 m/s in total time 10s.....
so displacement = 5×10 =50m
...
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It is Velocity and Time graph so we can say that A body is suffering this much Displacement from initial to final position in time taken.
Answer is the Area under graph since Area under graph would be equal to Quantity on Y-axis × Quantity on X-Axis
Hence, the Area under graph would give you
*Velocity* × Time
We know that Velocity × Time = Displacement
So Area Under Graph is the Answer
Find Area of ∆ ABF = 5
Area of BCFG = 4×5 = 20
Area of CGIE = 4×5 = 20
Area of ∆ CDE = 1/2 × 4 × 5 = 10
=> Total Area = 5+20+20+10
=> Total Displacement = 55 m
_______________________________
HOPE OT HELPS :):):)
Answer is the Area under graph since Area under graph would be equal to Quantity on Y-axis × Quantity on X-Axis
Hence, the Area under graph would give you
*Velocity* × Time
We know that Velocity × Time = Displacement
So Area Under Graph is the Answer
Find Area of ∆ ABF = 5
Area of BCFG = 4×5 = 20
Area of CGIE = 4×5 = 20
Area of ∆ CDE = 1/2 × 4 × 5 = 10
=> Total Area = 5+20+20+10
=> Total Displacement = 55 m
_______________________________
HOPE OT HELPS :):):)
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