Question

A particle is moving in parabolic path y = 4x2 with
constant speed v, m/s. The acceleration of the
particle when it crosses the origin is​

Answer 1

Answer:

8(Vx)^2.

Explanation:

We take dy/dt which is velocity along y direction or (Vy).

We take dx/dt which is velocity along x- direction or (Vx).

now from the parabolic equation we get that y =4x^2. So, on differentiating we will get that dy/dt = 8xdx/dt. Again, Vy will be 8x(Vx) so again on differentiating we will get d(Vy)/dt = 8(Vx)dx/dt where we know that (Vx) is constant.

So, (ay) is 8(Vx)(Vx) where ay is the acceleration along y-axis.

So, acceleration is 8(Vx)^2.

Answer 2

Given that,

A particle is moving in parabolic path : $y=4x^2$ with constant speed v m/s.

To find,

The acceleration of the  particle when it crosses the origin.

Solution,

Velocity, $v=\dfrac{dy}{dt}$

Since, $y=4x^2$

So,

On differentiating we get :

$\dfrac{dy}{dt}=8x\dfrac{dx}{dt}$

$\dfrac{dy}{dt}=v_{y}=8x\dfrac{dx}{dt}\\\\v_{y}= 8x\dfrac{dx}{dt}\\\\v_{y}=8xv$ ....(2)

Now, acceleration of the particle is :

$a=\dfrac{dv}{dt}$ (rate of change of velocity)

Differentiate equation (2) we get :

$a=8v\dfrac{dx}{dt}\\\\a=8v\times v\\\\a=8v^2$

So, the acceleration of the particle when it crosses the origin is $8v^2\ m/s^2$.