a walks with some time t with velocity (v) due east and then he walks for same time t with velocity (v) due north.the average velocity of man is
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Average velocity = Displacement / Time
On solving east and north walk then you will get hypotenuse as displacement
DISTANCE = velocity × time
distance in east = vt
distance in north = vt
Then hypotenuse = √ (vt)^2 + (vt)^2
= √ v^2 t^2 + v^2 t^2
= √ ( 2 v square t square )
= √2 vt
DISPLACEMENT = √2 vt
TIME = t + t
AVERAGE VELOCITY = √2 vt / ( t + t )
ANSWER = AVERAGE VELOCITY = √2 vt / ( t + t )
HOPE IT HELPS :):):):):):):):)
MARK AS BRAINLIEST
On solving east and north walk then you will get hypotenuse as displacement
DISTANCE = velocity × time
distance in east = vt
distance in north = vt
Then hypotenuse = √ (vt)^2 + (vt)^2
= √ v^2 t^2 + v^2 t^2
= √ ( 2 v square t square )
= √2 vt
DISPLACEMENT = √2 vt
TIME = t + t
AVERAGE VELOCITY = √2 vt / ( t + t )
ANSWER = AVERAGE VELOCITY = √2 vt / ( t + t )
HOPE IT HELPS :):):):):):):):)
MARK AS BRAINLIEST
BRAINLYY:
its an easier method
how can distance be velocity x time? its just a dought
velocity and speed are equal in linear motion
here the linear motion is in direction of east and then north
done
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