Question

A particle moves along the parabolic path x = y² + 2y + 2 in such a way that the Y component of the velocity vector remains 5m/s during the motion. The magnitude of the acceleration of the particle is

a) 50m/s²

b) 100m/s²

c) 10√2m/s²

d) 0.1m/s²​

Answer 1

differentiating x , we get v :

=>Vx = 2y Vy+ 2

=> Vx = 10y + 2

differentiating Vx , we get a :

=> a = 10 Vy

=> a = 10*5

=> a = 50 m s^-2

hope it helps

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Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

$\large{\bold{x = y^2 + 2y + 2}}$

Let the y - component of velocity be,

$\large{v_y = 5 \: m/s}$

$\huge{\bold{\underline{Explanation:-}}}$

Given equation,

$\large{x = y^2 + 2y + 2}$

Differentiating it,

$\large{ \dfrac{dx}{dt} = \dfrac{d(y^2)}{dt} + \dfrac{d(2y)}{dt} + 2}$

$\large{ \dfrac{dx}{dt} = 2y \times \dfrac{dy}{dt} + 2 \dfrac{dy}{dt} + 0}$

As we know,

dx/dt = velocity component of x.

and,

dy/dt = velocity component of y.

Substituting the values,

$\large{v_x = 2y \times v_y + 2 \times v_y}$

As we know, $\large{v_y = 5 \: m/s}$

Now,

$\large{v_x = 2y \times 5 + 2 \times 5}$

$\large{\boxed{v_x = 10y + 10}}$

Differentiating w.r.t time to get acceleration,

$\large{v_x = 10y + 10}$

$\large{ \dfrac{d v_x}{dt} = \dfrac{d(10y)}{dt} + 0}$

$\large{ \dfrac{d v_x}{dt} = 10 \times \dfrac{dy}{dt}}$

As we know,

dVx/dt = acceleration component of x.

component of x. and,

component of x. and,dy/dt = velocity component of y.

$\large{a_x = 10 \times v_y}$

As we know,

$\large{v_y = 5 \: m/s}$

Substituting the values,

$\large{a_x = 10 \times 5}$

$\large{\boxed{a_x = 50 \: m/s^2}}$

As we know,

Magnitude of acceleration is given by:-

$\large{\bold{a = \sqrt{(a_x)^2 + (a_y)^2}}}$

But as we know,

the Y component velocity is constant.

then differentiating it will give a value of Zero.

Therefore,

$\large{a = \sqrt{(50)^2 + 0}}$

$\huge{\boxed{\boxed{a = 50 \: m/s^2}}}$

So, the acceleration of the particle is 50 m/s² (Option - (a)).