Let's break down the problem and solve it step by step: ### **Given:** - The helicopter starts from rest at A. - Travels with constant acceleration \( a \). - Path AB is straight. - Speed at B, \( v_B = 79 \) m/s. - Distance AB: \( s_{AB} = \sqrt{(490^2 + 225^2)} = 543.4 \) m. - At the instant when \( h = 110 \) m (vertical height from the ground). We are to find at that instant: - \( \dot{R} \) (rate of change of distance from O to the helicopter) - \( \dot{\theta} \), \( \dot{\phi} \) (rates of change of the angles as measured by the radar at O). --- ### **1. Find the Acceleration, \( a \):** Using kinematics: \[ v_B^2 = v_A^2 + 2as_{AB} \] \[ 79^2 = 0^2 + 2a \cdot 543.4 \] \[ a = \frac{79^2}{2 \times 543.4} = \frac{6241}{1086.8} = 5.74 \ \text{m/s}^2 \] --- ### **2. Find the Helicopter's Speed When \( h = 110 \) m:** Let the distance from A to C along the path be \( s \). Since acceleration is constant: \[ v^2 = 2as \] Let \( s_C \) be the distance from A to C. #### Find \( s_C \) corresponding to \( h = 110 \): From the diagram: - The path AB is in the \( x-y \) plane. - \( \frac{h}{s} = \frac{225}{543.4} \Rightarrow \frac{110}{s_C} = \frac{225}{543.4} \) \[ s_C = \frac{110 \cdot 543.4}{225} = 265.64 \ \text{m} \] #### Now, find the speed at C: \[ v_C^2 = 2a s_C = 2 \times 5.74 \times 265.64 = 3048.3 \] \[ v_C = \sqrt{3048.3} = 55.23 \ \text{m/s} \] --- ### **3. Find Coordinates of C:** - \( x_C = \frac{490}{543.4} \cdot s_C = \frac{490}{543.4} \cdot 265.64 = 239.6 \) m - \( y_C = \frac{225}{543.4} \cdot s_C = \frac{225}{543.4} \cdot 265.64 = 110 \) m (matches \( h \)) - \( z_C = 110 \) m --- ### **4. Find \( R \) (Distance from O to C):** From the diagram: O is at (0, 0, 0), C is at (239.6, 110, 110) \[ R = \sqrt{x_C^2 + y_C^2 + z_C^2} = \sqrt{239.6^2 + 110^2 + 110^2} \] \[ R = \sqrt{57408.16 + 12100 + 12100} = \sqrt{81608.16} = 285.7\, \text{m} \] --- ### **5. Find \( \dot{R} \):** The velocity vector along the path from A to B: - Direction cosines: \( \frac{x_C}{s_C}, \frac{y_C}{s_C}, \frac{z_C}{s_C} \) - \( \vec{v} = v_C \cdot \left(\frac{x_C}{s_C}\hat{i} + \frac{y_C}{s_C}\hat{j} + \frac{z_C}{s_C}\hat{k}\right) \) \[ \vec{R} = (x_C, y_C, z_C), \quad \hat{R} = \frac{\vec{R}}{R} \] \[ \dot{R} = \vec{v} \cdot \hat{R} \] So, \[ \hat{R} = \left(\frac{239.6}{285.7}, \frac{110}{285.7}, \frac{110}{285.7}\right) \] \[ \vec{v} = 55.23 \cdot \left(\frac{239.6}{265.64}, \frac{110}{265.64}, \frac{110}{265.64}\right) \] Dot product: \[ \dot{R} = 55.23 \left[ \frac{239.6}{265.64} \cdot \frac{239.6}{285.7} + \frac{110}{265.64} \cdot \frac{110}{285.7} + \frac{110}{265.64} \cdot \frac{110}{285.7} \right] \] Calculate each term: - \( \frac{239.6}{265.64} \approx 0.9023 \) - \( \frac{239.6}{285.7} \approx 0.8386 \) - \( \frac{110}{265.64} \approx 0.4141 \) - \( \frac{110}{285.7} \approx 0.3849 \) So, - First term: \( 0.9023 \times 0.8386 = 0.7568 \) - Second term: \( 0.4141 \times 0.3849 = 0.1595 \) - Third term: \( 0.4141 \times 0.3849 = 0.1595 \) - Sum: \( 0.7568 + 0.1595 + 0.1595 = 1.0758 \) So, \[ \dot{R} = 55.23 \times 1.0758 = 59.43 \,\text{m/s} \] --- ### **6. Find \( \dot{\theta} \) and \( \dot{\phi} \):** #### **a. \( \dot{\theta} \):** - \( \tan\theta = \frac{y_C}{x_C} = \frac{110}{239.6} = 0.459 \) - \( \theta = \tan^{-1}(0.459) = 24.7^\circ \) \[ \dot{\theta} = \frac{d}{dt} \tan^{-1}\left(\frac{y}{x}\right) = \frac{1}{1 + (y/x)^2} \cdot \frac{x \dot{y} - y \dot{x}}{x^2} \] - \( \dot{x} = v_C \cdot \frac{x_C}{s_C} = 55.23 \cdot 0.9023 = 49.85 \) m/s - \( \dot{y} = v_C \cdot \frac{y_C}{s_C} = 55.23 \cdot 0.4141 = 22.88 \) m/s \[ \dot{\theta} = \frac{1}{1 + 0.459^2} \cdot \frac{239.6 \cdot 22.88 - 110 \cdot 49.85}{(239.6)^2} \] Calculate numerator: - \( 239.6 \cdot 22.88 = 5483.65 \) - \( 110 \cdot 49.85 = 5483.5 \) - \( 5483.65 - 5483.5 = 0.15 \) Denominator: - \( 1 + 0.2107 = 1.2107 \) - \( (239.6)^2 = 57408.16 \) So, \[ \dot{\theta} = \frac{1}{1.2107} \cdot \frac{0.15}{57408.16} = \frac{0.826}{57408.16} \approx 1.44 \times 10^{-5} \,\text{rad/s} \] --- #### **b. \( \dot{\phi} \):** - \( \tan\phi = \frac{z_C}{\sqrt{x_C^2 + y_C^2}} = \frac{110}{\sqrt{239.6^2 + 110^2}} = \frac{110}{263.45} = 0.4177 \) - \( \phi = \tan^{-1}(0.4177) = 22.7^\circ \) \[ \dot{\phi} = \frac{\sqrt{x^2 + y^2} \dot{z} - z \frac{x \dot{x} + y \dot{y}}{\sqrt{x^2 + y^2}}}{x^2 + y^2 + z^2} \] - \( \sqrt{x_C^2 + y_C^2} = 263.45 \) - \( x_C \dot{x} + y_C \dot{y} = 239.6 \cdot 49.85 + 110 \cdot 22.88 = 11945.06 + 2516.8 = 14461.86 \) - \( \dot{z} = v_C \cdot \frac{z_C}{s_C} = 55.23 \cdot 0.4141 = 22.88 \) m/s Numerator: \[ 263.45 \cdot 22.88 - 110 \cdot \frac{14461.86}{263.45} = 6028.5 - 110 \times 54.92 = 6028.5 - 6041.2 = -12.7 \] Denominator: \[ x^2 + y^2 + z^2 = 57408.16 + 12100 + 12100 = 81608.16 \] So, \[ \dot{\phi} = \frac{-12.7}{81608.16} = -1.56 \times 10^{-4} \ \text{rad/s} \] --- ## **Summary of Answers** - \( \dot{R} = 59.43 \) m/s - \( \dot{\theta} = 1.44 \times 10^{-5} \) rad/s - \( \dot{\phi} = -1.56 \times 10^{-4} \) rad/s Let me know if you'd like to see the step-by-step breakdown for any part!