Suppose the position vector for a particle is given as a function of time by \( \vec{r}(t)=(t+1) \hat{\imath}+\left(0.125 t^{2}+1\right) \hat{\jmath} \) Determine the particle's speed at \( t=2.00 \mathrm{~s} \).

To determine the particle's speed at \( t=2.00 \mathrm{~s} \), we first need to find the velocity vector by differentiating the position vector \( \vec{r}(t) \). The position vector is given by: \[ \vec{r}(t)=(t+1) \hat{\imath}+\left(0.125 t^{2}+1\right) \hat{\jmath} \] Now, let's differentiate with respect to time \( t \): \[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}[(t+1) \hat{\imath} + (0.125 t^2 + 1) \hat{\jmath}] \] \[ \vec{v}(t) = \hat{\imath} + (0.25 t) \hat{\jmath} \] Now substituting \( t = 2.00 \mathrm{~s} \): \[ \vec{v}(2) = \hat{\imath} + (0.25 \cdot 2) \hat{\jmath} = \hat{\imath} + 0.5 \hat{\jmath} \] Next, we find the speed by calculating the magnitude of the velocity vector: \[ |\vec{v}(2)| = \sqrt{(1)^2 + (0.5)^2} = \sqrt{1 + 0.25} = \sqrt{1.25} \approx 1.414 \text{ m/s} \] Thus, the particle's speed at \( t=2.00 \mathrm{~s} \) is approximately \( 1.41 \text{ m/s} \).