QJEEM23Ph4e32: JEE Main 2023 Physics - Class 11 JEEM23Ph4e32 Solution: The position of a particle related to time is given by . The…

The position of a particle as a function of time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$ .

To find the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ , we first need to find the velocity of the particle as a function of time.

The velocity $v$ is the time derivative of the position $x$ :

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

Taking the derivative of $x$ with respect to $t$ , we get:

$v = \frac{dx}{dt} = 10t - 4\,\mathrm{m/s}$

Now we can find the velocity of the particle at $t=2\,\mathrm{s}$ by plugging in $t=2$ :

$v(2\,\mathrm{s}) = 10(2) - 4\,\mathrm{m/s} = 16\,\mathrm{m/s}$

Therefore, the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ is:

$\boxed{|v(2\,\mathrm{s})| = 16\,\mathrm{m/s}}$