Question

A particle moves along a path x=y^2+2y+2 in a plane such that y-component of its velocity remains constant at 2 m/s.the magnitude of its acceleration is​

Answer 1

Given:

A particle moves along a path x=y^2+2y+2 in a plane such that y-component of its velocity remains constant at 2 m/s.

To find:

The magnitude of Acceleration of particle ?

Calculation:

Acceleration can be calculated by the 2nd order

derivative of displacement function; since the y component of velocity is constant , the particle will have acceleration only along x axis :

$\therefore \: x = {y}^{2} + 2y + 2$

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

$\implies \: \dfrac{dx}{dt} = 2y( \dfrac{dy}{dt} ) + 2( \dfrac{dy}{dt} ) + 0$

Putting dy/dt = 2 (as given in question)

$\implies \: \dfrac{dx}{dt} = 2y(2) + 2(2)$

$\implies \: \dfrac{dx}{dt} = 4y + 4$

$\implies \: \dfrac{ {d}^{2}x }{d {t}^{2} } = \dfrac{d(4y + 4)}{dt}$

$\implies \: a = 4( \dfrac{dy}{dt} ) + 0$

$\implies \: a = 4(2)$

$\implies \: a =8 \: m/ {s}^{2}$

So, acceleration of particle is 8 m/s².