Question

a particle is moving in parabolic path y = 4 x square with constant speed V metre per second the acceleration of the particle when it crosses the origin is​


first answer will be the brainlist

Answer 1

Answer:8V^2 m/s^2

Explanation:

Answer 2

Answer:

The acceleration of the particle is 8v² m/s².

Explanation:

Given that,

$y=4x^2$

Speed = v m/s

Velocity :

The velocity is the first derivative of the position of the particle.

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

We need to calculate the velocity

The position of the particle

$y=4x^2$....(I)

On differentiating equation (I)

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

The velocity is

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

$v_{y}= 8x\dfrac{dx}{dt}$

$v_{y}=8xv$...(II)

We need to calculate the acceleration

Acceleration :

The acceleration is the first derivative of the velocity of the particle.

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

Now, differentiating equation (II)

$a=8v\dfrac{dx}{dt}$

$a=8v\times v$

$a=8v^2$

Hence, The acceleration of the particle is 8v² m/s².