Let's break down the problem step by step. ## **Given Data** - **Weight of particle (W):** 10 lb - **Forces:** - \(\vec{F}_1 = (t^2 \hat{i} + 3\hat{j} - 4t\hat{k}) \) lb - \(\vec{F}_2 = (1 - 4t)\hat{i} \) lb - \(\vec{F}_3 = (2t^2 \hat{i} - t\hat{j} - t\hat{k}) \) lb - **Released from rest at** \( t = \) - **Find:** Distance from origin at \( t = 3 \) s --- ## **Step 1: Find the Mass** Weight \( W = mg \Rightarrow m = \frac{W}{g} \) - \( g = 32.2 \) ft/s² (standard gravity) - \( m = \frac{10}{32.2} = .3106 \) slugs --- ## **Step 2: Write the Total Force** Sum all forces, including weight (downward): \[ \vec{W} = -10\hat{k} \text{ lb} \] Sum up: \[ \vec{F}_{\text{total}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{W} \] Calculate each component: - **\(\hat{i}\) component:** \[ (t^2) + (1 - 4t) + 2t^2 = 3t^2 - 4t + 1 \] - **\(\hat{j}\) component:** \[ 3 + (-t) = 3 - t \] - **\(\hat{k}\) component:** \[ -4t + (-t) - 10 = -5t - 10 \] So, \[ \vec{F}_{\text{total}} = (3t^2 - 4t + 1)\hat{i} + (3 - t)\hat{j} + (-5t - 10)\hat{k} \text{ lb} \] --- ## **Step 3: Find Acceleration** \[ \vec{a} = \frac{\vec{F}_{\text{total}}}{m} \] Where \( m = .3106 \) slugs. \[ \vec{a}(t) = \frac{(3t^2 - 4t + 1)\hat{i} + (3 - t)\hat{j} + (-5t - 10)\hat{k}}{.3106} \text{ ft/s}^2 \] --- ## **Step 4: Find Velocity by Integrating Acceleration** Since released from rest, \( \vec{v}() = \). Integrate each component from \( \) to \( t \): - **\(\hat{i}\):** \[ \int_^t \frac{3\tau^2 - 4\tau + 1}{.3106} d\tau = \frac{1}{.3106} \left[ t^3 - 2t^2 + t \right] \] - **\(\hat{j}\):** \[ \int_^t \frac{3 - \tau}{.3106} d\tau = \frac{1}{.3106} (3t - \frac{1}{2} t^2) \] - **\(\hat{k}\):** \[ \int_^t \frac{-5\tau - 10}{.3106} d\tau = \frac{1}{.3106} \left[ -\frac{5}{2} t^2 - 10t \right] \] --- ## **Step 5: Find Position by Integrating Velocity** Integrate each velocity component from \( \) to \( t \): - **\(\hat{i}\):** \[ \int_^t \frac{t^3 - 2t^2 + t}{.3106} dt = \frac{1}{.3106} \left[ \frac{1}{4} t^4 - \frac{2}{3} t^3 + \frac{1}{2} t^2 \right] \] - **\(\hat{j}\):** \[ \int_^t \frac{3t - \frac{1}{2} t^2}{.3106} dt = \frac{1}{.3106} \left[ \frac{3}{2} t^2 - \frac{1}{6} t^3 \right] \] - **\(\hat{k}\):** \[ \int_^t \frac{-\frac{5}{2} t^2 - 10t}{.3106} dt = \frac{1}{.3106} \left[ -\frac{5}{6} t^3 - 5 t^2 \right] \] --- ## **Step 6: Plug in \( t = 3 \) s** Calculate each component: ### **\(\hat{i}\) Component** \[ x = \frac{1}{.3106} \left[ \frac{1}{4} (3)^4 - \frac{2}{3} (3)^3 + \frac{1}{2} (3)^2 \right] \] \[ = \frac{1}{.3106} \left[ \frac{1}{4}(81) - \frac{2}{3}(27) + \frac{1}{2}(9) \right] \] \[ = \frac{1}{.3106} \left[ 20.25 - 18 + 4.5 \right] = \frac{1}{.3106}(6.75) \approx 21.74 \text{ ft} \] ### **\(\hat{j}\) Component** \[ y = \frac{1}{.3106} \left[ \frac{3}{2} (9) - \frac{1}{6} (27) \right] = \frac{1}{.3106} [13.5 - 4.5] = \frac{1}{.3106}(9) \approx 28.99 \text{ ft} \] ### **\(\hat{k}\) Component** \[ z = \frac{1}{.3106} \left[ -\frac{5}{6} (27) - 5 (9) \right] = \frac{1}{.3106} [ -22.5 - 45 ] = \frac{1}{.3106}(-67.5) \approx -217.41 \text{ ft} \] --- ## **Step 7: Distance from Origin** \[ d = \sqrt{ x^2 + y^2 + z^2 } \] \[ d = \sqrt{ (21.74)^2 + (28.99)^2 + (-217.41)^2 } \] \[ = \sqrt{ 472.89 + 840.41 + 47282.67 } \] \[ = \sqrt{ 48596 } \] \[ \approx 220.44 \text{ ft} \] --- ## **Final Answer** \[ \boxed{220 \text{ ft}} \] **The particle is approximately 220 ft from the origin 3 seconds after being released from rest.**