91Ó°ÊÓ

The position vector \( \vec{r}(t) \) of the dot is given by: \[ \vec{r} = [34.0 \text{ cm } +(2.5 \text{ cm/s}^2)t^2] \hat{i} +(5.0 \text{ cm/s})t \hat{\jmath}. \] We need to find the average velocity for \( t = 0 \text{ s} \) to \( t = 2.0 \text{ s} \), and the instantaneous velocities at specific moments \( t = 0, 1.0, 2.0 \text{ s}. \)

The average velocity \( \vec{v}_{\text{avg}} \) is the change in position divided by the change in time: \[ \vec{v}_{\text{avg}} = \frac{\vec{r}(2.0) - \vec{r}(0)}{2.0 - 0}. \] Substitute \( t = 2.0 \text{ s} \) and \( t = 0 \text{ s} \) into the position function to find \( \vec{r}(2.0) \) and \( \vec{r}(0) \). For \( t = 0: \) \[ \vec{r}(0) = [34.0] \hat{i} + [0] \hat{\jmath} = 34.0 \hat{i} \text{ cm}. \] For \( t = 2.0: \)\[ \vec{r}(2.0) = [34.0 + 2.5 \times (2.0)^2] \hat{i} + [5.0 \times 2.0] \hat{\jmath}. \] \[ = [44.0] \hat{i} + [10.0] \hat{\jmath} \text{ cm}. \] Therefore, \[ \vec{v}_{\text{avg}} = \frac{(44.0 - 34.0) \hat{i} + (10.0 - 0) \hat{\jmath}}{2.0}. \] \[ = 5.0 \hat{i} + 5.0 \hat{\jmath} \text{ cm/s}. \] The magnitude of \( \vec{v}_{\text{avg}}\) is found by: \[ \|\vec{v}_{\text{avg}}\| = \sqrt{(5.0)^2 + (5.0)^2} = \sqrt{50} \approx 7.1 \text{ cm/s}. \] The direction is given by: \[ \theta = \tan^{-1}\left(\frac{5.0}{5.0}\right) = 45^\circ \text{ with respect to the } \hat{i}\text{-axis}. \]

The instantaneous velocity \( \vec{v} \) is the derivative of the position vector \( \vec{r} \) with respect to \( t \). Differentiate \( \vec{r}(t) = [34.0 + 2.5t^2] \hat{i} + [5.0t] \hat{\jmath} \) with respect to \( t \): \[ \vec{v}(t) = \frac{d}{dt}([34.0 + 2.5t^2] \hat{i}) + \frac{d}{dt}([5.0t] \hat{\jmath}). \] \[ = [5.0t] \hat{i} + [5.0] \hat{\jmath}. \] Now calculate \( \vec{v}(t) \) for \( t = 0, 1.0, 2.0 \text{ s}. \)At \( t = 0 \text{ s}: \) \[ \vec{v}(0) = [5.0 \times 0] \hat{i} + [5.0] \hat{\jmath} = 0 \hat{i} + 5.0 \hat{\jmath} \text{ cm/s}. \]The magnitude is \[\|\vec{v}(0)\| = 5.0 \text{ cm/s}. \] and direction is \( 90^\circ \).At \( t = 1.0 \text{ s}: \) \[ \vec{v}(1.0) = [5.0 \times 1.0] \hat{i} + [5.0] \hat{\jmath} = 5.0 \hat{i} + 5.0 \hat{\jmath} \text{ cm/s}. \] The magnitude is \( \|\vec{v}(1.0)\| \approx 7.1 \text{ cm/s} \) and the direction is \( 45^\circ \).At \( t = 2.0 \text{ s}: \) \[ \vec{v}(2.0) = [5.0 \times 2.0] \hat{i} + [5.0] \hat{\jmath} = 10.0 \hat{i} + 5.0 \hat{\jmath} \text{ cm/s}. \] The magnitude is \( \|\vec{v}(2.0)\| \approx 11.2 \text{ cm/s} \) and the direction is \( \tan^{-1}(0.5) \approx 26.6^\circ \).

Plot the position vector \( \vec{r}(t) \) at several points between \( t = 0 \text{ s} \) and \( t = 2.0 \text{ s} \). Notice the path starts from \( 34.0 \hat{i} \text{ cm} \) and moves in an arc towards \( 44.0 \hat{i} + 10.0 \hat{\jmath} \text{ cm} \). Along this path, show the calculated velocities: - At \( t = 0 \text{ s} \), vector pointing vertically upwards, - At \( t = 1.0 \text{ s} \), vector at \( 45^\circ \), - At \( t = 2.0 \text{ s} \), vector closer to the \( \hat{i} \) due to larger \( \hat{i} \) component.