Question

13.
The position (x) of a particle moving along x-axis, is
given by x = 12t2 - 413, where x is in metre and time
t is in second. The position of the particle when it
achieves maximum speed along positive
X-direction, is
Ti 16 m
(2) 4 m
1378 m
(4) 2 m​

Answer 1

The position of the particle when it  achieves maximum speed along positive  x-direction is 8 meters.

Explanation:

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

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

Where

x is in meter

t is in seconds

We know that velocity of a particle is given by :

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

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

$v=(24t-12t^2)\ m/s$............(1)

We need to find the position of the particle when it  achieves maximum speed along positive  x-direction :

For maximum speed,

$\dfrac{dv}{dt}=0$

$\dfrac{d(24t-12t^2)}{dt}=0$

t = 1 s

The position of the particle at t = 1 s is :

$x=12(1)^2-4(1)^3$

x = 8 meters

So, the position of the particle when it  achieves maximum speed along positive  x-direction is 8 meters. Hence, this is the required solution.

Learn more,

Maxima and minima

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Answer 2

and i am sure that i am not going to change my habits tq