The position of a particle moving along the \(x\) -axis varies with time according to the expression \(x=4 t^{2},\) where \(x\) is in meters and \(t\) is in seconds. Evaluate the particle's position a) at \(t=2.00 \mathrm{~s}\). b) at \(2.00 \mathrm{~s}+\Delta t\) c) Evaluate the limit of \(\Delta x / \Delta t\) as \(\Delta t\) approaches zero, to find the velocity at \(t=2.00 \mathrm{~s}\).

To find the position of the particle at \(t = 2.00\) s, plug the value of \(t\) into the given expression for position \(x\): \(x = 4t^2\) \(x(2.00) = 4(2.00)^2\) \(x(2.00) = 4(4)\) \(x(2.00) = 16 \, \text{m}\) The particle's position at \(t = 2.00\) s is 16 meters.

To find the position at \(t = 2.00 \, \text{s} + \Delta t\), plug this value of \(t\) into the position function: \(x(t) = 4t^2\) \(x(2.00+\Delta t) = 4(2.00+\Delta t)^2\) The position of the particle at \(t = 2.00\,\text{s} + \Delta t\) is \(x(2.00+\Delta t) = 4(2.00+\Delta t)^2\).

To find the velocity of the particle at \(t = 2.00\) s, we need to find the limit of the difference quotient as \(\Delta t\) approaches zero: \(\lim_{\Delta t \to 0} \frac{x(2.00+\Delta t) - x(2.00)}{\Delta t}\) First, let's simplify the expression numerator: \(x(2.00+\Delta t) - x(2.00) = 4(2.00+\Delta t)^2 - 4(2.00)^2\) \(= 4[(2.00+\Delta t)^2 - (2.00)^2]\) Next, factor the difference of squares in the brackets: \(= 4[(2.00+\Delta t - 2.00)(2.00+\Delta t + 2.00)]\) \(= 4[\Delta t(4.00+\Delta t)]\) Now, let's put this back into our limit: \(\lim_{\Delta t \to 0} \frac{4[\Delta t(4.00+\Delta t)]}{\Delta t}\) We can cancel the \(\Delta t\) terms: \(\lim_{\Delta t \to 0} 4(4.00+\Delta t)\) Now, take the limit as \(\Delta t\) approaches zero: \(4(4.00 + 0)\) \(= 4(4.00)\) \(= 16 \, \mathrm{m/s}\) The particle's velocity at \(t = 2.00\) s is 16 m/s.