Question

the position(x) of a particle moving along x axis is given by x=12t^2-4t^2. the position of the particle when it achieves maximum speed along positive x direction,is​

Answer 1

The position of the particle when it achieves maximum speed is 2 meters.

Explanation:

The position of a particle moving along x axis is given by :

$x=12t^2-4t^3$

Where

t is time in seconds

Let v is the speed of the particle. The velocity can be calculated as :

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

$v=\dfrac{d(12t^2-4t^3)}{dt}$

$v=24t-12t^2$

For maximum speed,

v = 0

$24t-12t^2=0$

x = 2 m

So, the the position of the particle when it achieves maximum speed along positive x direction is 2 meters. Hence, this is the required solution.

Learn more :

Topic : Speed and displacement

https://brainly.in/question/5382706