Certainly! Here’s the transcribed problem and a step-by-step solution: --- ### **Problem** An object moves in a plane. Its position at time \( t \) (in seconds) is described by: \[ \vec{r}(t) = (4t^2 \hat{i} + 3t \hat{j}) \ \text{m} \] (a) Find the object's velocity and acceleration as functions of time. (b) At time \( t = 1. \) s, determine the magnitude and direction (relative to the x-axis) of its velocity. (c) Your answer should be correct to two significant figures. --- ## **Step-by-Step Solution** --- ### **(a) Velocity and Acceleration as Functions of Time** #### **Velocity** Velocity is the time derivative of position: \[ \vec{v}(t) = \frac{d\vec{r}}{dt} \] Given: \[ \vec{r}(t) = 4t^2 \hat{i} + 3t \hat{j} \] Differentiate each component: - \( \frac{d}{dt}(4t^2) = 8t \) - \( \frac{d}{dt}(3t) = 3 \) So: \[ \vec{v}(t) = 8t \hat{i} + 3 \hat{j} \] #### **Acceleration** Acceleration is the time derivative of velocity: \[ \vec{a}(t) = \frac{d\vec{v}}{dt} \] Differentiate each component: - \( \frac{d}{dt}(8t) = 8 \) - \( \frac{d}{dt}(3) = \) So: \[ \vec{a}(t) = 8 \hat{i} \] --- ### **(b) Magnitude and Direction of Velocity at \( t = 1. \) s** Plug \( t = 1. \) s into the velocity: \[ \vec{v}(1.) = 8 \cdot 1. \hat{i} + 3 \hat{j} = 8\hat{i} + 3\hat{j} \] #### **Magnitude:** \[ |\vec{v}| = \sqrt{8^2 + 3^2} = \sqrt{64 + 9} = \sqrt{73} \approx 8.5 \ \text{m/s} \] #### **Direction (relative to x-axis):** The direction \( \theta \) is: \[ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{3}{8}\right) \] Calculate: \[ \theta \approx \tan^{-1}(.375) \approx 21^\circ \] --- ## **Final Answers** ### (a) - Velocity as a function of time: \(\boxed{\vec{v}(t) = 8t \hat{i} + 3\hat{j}}\) - Acceleration as a function of time: \(\boxed{\vec{a}(t) = 8\hat{i}}\) ### (b) At \( t = 1. \) s: - Magnitude of velocity: \(\boxed{8.5\ \text{m/s}}\) - Direction relative to x-axis: \(\boxed{21^\circ}\)