The position of a particle related to time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$. The magnitude of velocity of the particle at $t=2 s$ will be :

The position of a particle as a function of time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$. <br/><br/>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. <br/><br/> The velocity $v$ is the time derivative of the position $x$: <br/><br/> $v = \frac{dx}{dt}$ <br/><br/> Taking the derivative of $x$ with respect to $t$, we get: <br/><br/> $v = \frac{dx}{dt} = 10t - 4\,\mathrm{m/s}$ <br/><br/> Now we can find the velocity of the particle at $t=2\,\mathrm{s}$ by plugging in $t=2$: <br/><br/> $v(2\,\mathrm{s}) = 10(2) - 4\,\mathrm{m/s} = 16\,\mathrm{m/s}$ <br/><br/> Therefore, the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ is: <br/><br/> $\boxed{|v(2\,\mathrm{s})| = 16\,\mathrm{m/s}}$