Suppore the position veclor for a particle is given as a function of tirne by \( \left.\vec{r}(t)=(t+1) t+\left(0.125 t^{2}+1\right)\right) \) Determine the particle's speed at \( t=2.00 \) s

It seems there was an issue with calculating the derivative. Let's try to calculate it manually. The derivative of the position vector with respect to time \( t \) is given by: \[ \vec{v}(t) = \frac{d\vec{r}(t)}{dt} \] Let's calculate the derivative manually: \[ \vec{v}(t) = \frac{d}{dt}[(t+1)t + (0.125t^2 + 1)] \] \[ \vec{v}(t) = \frac{d}{dt}[t^2 + t + 0.125t^2 + 1] \] \[ \vec{v}(t) = \frac{d}{dt}[1.125t^2 + t + 1] \] \[ \vec{v}(t) = 2.25t + 1 \] Now, we have the velocity vector as a function of time \( t \). To find the particle's speed at \( t = 2.00 \) s, we need to find the magnitude of the velocity vector at that time. The magnitude of the velocity vector is given by the square root of the sum of the squares of its components. Let's calculate the magnitude of the velocity vector at \( t = 2.00 \) s. Calculate the value by following steps: - step0: Calculate: \(\sqrt{\left(2.25\times 2+1\right)^{2}}\) - step1: Multiply the numbers: \(\sqrt{\left(4.5+1\right)^{2}}\) - step2: Add the numbers: \(\sqrt{5.5^{2}}\) - step3: Simplify the root: \(5.5\) The particle's speed at \( t = 2.00 \) s is \( 5.5 \) m/s.