Question

A particle moves along the parabola y=x2 such that x component of velocity is always 2m/s. The acceleration of the particle is

Answer 1

Answer:

y= x^2

dy/dt = d/dt(x^2)

Vy=2x*dx/dt

Vy = 2xVx

Vy= 2*x*2 = 4x

ay= dVy/dt = 4*Vx= 8m/s^2

ax=d/dt(Vx) = d/dt(2) = 0

Therefore net acceleration is 8m/s^2

Answer 2

Answer:

The acceleration of the particle is $8m/s^2$.

Explanation:

Given the parabolic path of a particle,  $y=x^{2}$

Velocity of x component, $v_x= 2m/s$

Consider the path of the particle,

$y=x^{2}$

Differentiating both sides with respect to t,

$\dfrac{dy}{dt} = \dfrac{dx^{2} }{dt}$

$\implies \dfrac{dy}{dt} =2x \dfrac{dx }{dt}$

Using the definition of velocity, rate of change of displacement is called velocity. Let us take displacement along x-component as $v_x$ and $v_y$ along y-component.

Thus, the above equation can be written as

$v_y = 2xv_x$

Substituting value of $v_x$

$\implies v_y = 2x\times 2=4x$

Acceleration is defined as rate of change of velocity, and is given by

$a= \dfrac{dv_y}{dt}$

As x component of velocity is constant, acceleration has only y-component. Substituting $v_{y}$,

$a= \dfrac{d(4x)}{dt}=4\dfrac{dx}{dt}$

$=4v_x=4\times 2=8m/s^{2}$

Therefore, the net acceleration of the particle is $8m/s^2$.