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anmol jain Grade: 11
        A particle mover along the parabolic path x=y^2 + 2y + 2 in such a way that the y-component of velocity  vector remains 5 m/s during the motion.The magnitude of acceleration of the particle is-
7 years ago

Answers : (1)

pratham ashish
9 Points
										Hi,


A = root{(Ax)^2 + (Ay)^2}

A= total acceleration

Ax = acceleration along x axis = d^2x/dt^2
Ay = acceleration along y axis = d^2y/dt^2

here vel. in y directn =dy/dt = 5 m/s (constant) => Ay = 0

now differentiate the given eqn


=> Vx = 2y*dy/dt +2* dy/dt
= 2y*5 + 2 *5
Vx = 10y + 10

again diff.

Ax = 10 *dy/dt = 10* 5 = 50 m/s^2

A = root{(Ax)^2 + (Ay)^2} => A = Ax = 50 (as Ay=0 )


7 years ago
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