Question

A particle moves along the parabolic path y = 2x --X2 + 2. in such a way that the x-component of Veloch
vector remains constant (5 m/s). Find the magnitude of acceleration of the particle.​

Answer 1

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From the Question,

• Horizontal Velocity of the particle = 5 m/s

The path of the object is defined by the relation:

$\sf{y = 2x - x {}^{2} + 2 }$

Differentiating the relation,we obtain:

$\sf{ \frac{dy}{dt} = \frac{d( - x {}^{2} + 2x + 2)}{dx} } \\ \\ \implies \: \sf{ \frac{dy}{dt} = - 2x . \frac{dx}{dt} + 2. \frac{dx}{dt} }$

Here,

dx/dt is the horizontal component of velocity,which is 5 m/s

Also,

dy/dt is the vertical component of velocity

Now,

$\implies \: \sf{ \frac{dy}{dt} = - 2x(5) + 2(5) }$

$\implies \huge{\sf{{V}_{y} = - 10 m{s}^{-1}}}$

Now,

• Acceleration along x - axis is 0 m/s²

Acceleration along y - axis:

$\sf{{a}_{y} = -10 \times {V}_{y}} \\ \\ \sf{{a}_{y} = -100 m{s}^{-2}}$

We know that,

$\huge{\sf{a = \sqrt{{{a}^{2}}_{x} + {{a}^{2}}_{y}}}}$

Putting the values,we get:

$\implies \: \sf{a = \sqrt{( 0) {}^{2} + ( - 50) {}^{2} } } \\ \\ \huge{ \implies \: \sf{a = 50ms {}^{ - 2} }}$

Thus,the acceleration of the particle is 50 m/s²

Answer 2

Hey mate

refer to the attachment..