The problem describes a radially slotted arm whose rotation is governed by $\theta = 0.2t + 0.02t^2$. The distance of a slider B from point O is given by $r = 0.2 + 0.04t^2$. We are asked to calculate the magnitudes of the velocity and acceleration of the slider at time $t = 3$ s. $\theta$ is in radians and $r$ is in meters.

1. Problem Description

The problem describes a radially slotted arm whose rotation is governed by $\theta = 0.2t + 0.02t^2$. The distance of a slider B from point O is given by $r = 0.2 + 0.04t^2$. We are asked to calculate the magnitudes of the velocity and acceleration of the slider at time $t = 3$ s. $\theta$ is in radians and $r$ is in meters.

2. Solution Steps

First, we calculate the first and second derivatives of $\theta$ and $r$ with respect to time.

$$\theta = 0.2t + 0.02t^2$$

$$\dot{\theta} = \frac{d\theta}{dt} = 0.2 + 0.04t$$

$$\ddot{\theta} = \frac{d^2\theta}{dt^2} = 0.04$$

$$r = 0.2 + 0.04t^2$$

$$\dot{r} = \frac{dr}{dt} = 0.08t$$

$$\ddot{r} = \frac{d^2r}{dt^2} = 0.08$$

Next, we evaluate these derivatives at $t = 3$ s.

$$\dot{\theta}(3) = 0.2 + 0.04(3) = 0.2 + 0.12 = 0.32 \text{ rad/s}$$

$$\ddot{\theta}(3) = 0.04 \text{ rad/s}^2$$

$$r(3) = 0.2 + 0.04(3^2) = 0.2 + 0.04(9) = 0.2 + 0.36 = 0.56 \text{ m}$$

$$\dot{r}(3) = 0.08(3) = 0.24 \text{ m/s}$$

$$\ddot{r}(3) = 0.08 \text{ m/s}^2$$

Now, we calculate the radial and transverse components of velocity and acceleration.

$$v_r = \dot{r}$$

$$v_\theta = r\dot{\theta}$$

$$a_r = \ddot{r} - r\dot{\theta}^2$$

$$a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}$$

At $t=3$ s,

$$v_r(3) = \dot{r}(3) = 0.24 \text{ m/s}$$

$$v_\theta(3) = r(3)\dot{\theta}(3) = 0.56(0.32) = 0.1792 \text{ m/s}$$

$$a_r(3) = \ddot{r}(3) - r(3)\dot{\theta}(3)^2 = 0.08 - 0.56(0.32)^2 = 0.08 - 0.56(0.1024) = 0.08 - 0.057344 = 0.022656 \text{ m/s}^2$$

$$a_\theta(3) = r(3)\ddot{\theta}(3) + 2\dot{r}(3)\dot{\theta}(3) = 0.56(0.04) + 2(0.24)(0.32) = 0.0224 + 0.1536 = 0.176 \text{ m/s}^2$$

Finally, we calculate the magnitudes of the velocity and acceleration.

$$v = \sqrt{v_r^2 + v_\theta^2} = \sqrt{(0.24)^2 + (0.1792)^2} = \sqrt{0.0576 + 0.03211264} = \sqrt{0.08971264} \approx 0.2995 \text{ m/s}$$

$$a = \sqrt{a_r^2 + a_\theta^2} = \sqrt{(0.022656)^2 + (0.176)^2} = \sqrt{0.0005133 + 0.030976} = \sqrt{0.0314893} \approx 0.1775 \text{ m/s}^2$$

3. Final Answer

The magnitude of the velocity of the slider at $t=3$ s is approximately $0.2995$ m/s.

The magnitude of the acceleration of the slider at $t=3$ s is approximately $0.1775$ m/s$^2$.