Question

A particle move along the parabolic path,
x = y2 + 2y + 2 in such a way that the
y-component of velocity vector remain 5 m/s during
the motion. The magnitude of the acceleration of
the particle is
(1) 50 m/s2
(2) 100 m/s2
(3) 10/2 m/s2
(4) 0.1 m/s2​

Answer 1

◆【SOLUTION】◆

HEY , WE CAN SOLVE SUCH TYPE OF QUESTIONS BY USING DIFFERENTIAL KNOWLEDGE.

●【DIFFERENTIAL NOTATIONS 】●

♀ Velocity = dy/dt

♀ Acceleration = d²x/dt²

◆【Function given】◆

X = y² + 2y + 2

Differentiate the equation

dx/dt = 2y(dy/dt) + 2(dy/dt) + 0

v(x) = 2y.v(y) +2.v(y)

where v(x) represent the velocity in x direction and v(y) represent the velocity in y direction.

v(y) = 5m/sec

v(x) = 10y + 20

again differentiate with respect to time.

dv/dt = 10(dy/dt) + 0

a(x) = 10×5

So acceleration in x direction is 50m/sec²

but the acceleration in y direction is zero because velocity is constant.

◆【|Total acceleration| = 50】◆

♀ Option A ♀

#answerwithquality

#BAL