A ball is moving along the hyperbolic path xy=9 with constant speed of 8sqrt(10)m/s .Velocity of particle (in m/s) along x -direction when it at x=sqrt(3) is
Answers
Given : A ball is moving along the hyperbolic path xy=9 with constant speed of 8√10 m/s
To Find : Velocity of particle (in m/s) along x -direction when it at x=√3 is
Solution :
xy = 9
differentiating wrt t
=> xdy/dt + ydx/dt = 0
=> dy/dt = -(y/x)dx/dt
dx/dt = horizontal Velocity - Velocity along x direction
dy/dt = vertical velocity - Velocity along y direction
Net velocity =√((dx/dt)² + (dy/dt)²)
Speed is constant = 8√10 m/s
Hence √((dx/dt)² + (dy/dt)²) = 8√10
Squaring both sides
=> (dx/dt)² + (dy/dt)² = 640
Substituting dy/dt
=> (dx/dt)² + ( -(y/x)dx/dt)² = 640
=> (dx/dt)²( 1 + (-y /x)²) = 640
=>
x = √3
xy = 9
y = 9/x = 9/√3 = 3√3
y/x = 3√3 / √3 = 3
=> (dx/dt)²( 1 + (-3)²) = 640
=> (dx/dt)²( 1 + 9) = 640
=> (dx/dt)²( 1 0) = 640
=> (dx/dt)² = 64
=> dx/dt = 8 m/s
Velocity of particle (in m/s) along x -direction when it at x=√3 ) is 8 m/s
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