Let me clarify the problem and restate the information: - The initial speed \( v_0 = 50 \) m/sec - The final speed \( v = 20 \) m/sec - The acceleration is given by: \[ a = -4 \times 10^{-3} v^2 \quad \text{(where \( v \) is in m/sec and \( a \) is in m/sec}^2\text{)} \] - We are to **determine the time elapsed** for the speed to decrease from 50 m/sec to 20 m/sec. Let's proceed step by step. --- ## 1. Write the acceleration as a differential equation Acceleration is the time derivative of velocity: \[ a = \frac{dv}{dt} = -4 \times 10^{-3} v^2 \] --- ## 2. Separate the variables \[ \frac{dv}{dt} = -4 \times 10^{-3} v^2 \] \[ \frac{dv}{v^2} = -4 \times 10^{-3} dt \] --- ## 3. Integrate both sides Integrate from \( v_0 = 50 \) to \( v = 20 \) and \( t = 0 \) to \( t = t_1 \): \[ \int_{50}^{20} v^{-2} dv = -4 \times 10^{-3} \int_{0}^{t_1} dt \] \[ \left[ -v^{-1} \right]_{50}^{20} = -4 \times 10^{-3} (t_1 - 0) \] \[ \left( -\frac{1}{20} + \frac{1}{50} \right) = -4 \times 10^{-3} t_1 \] --- ## 4. Simplify the numbers \[ -\frac{1}{20} + \frac{1}{50} = -\frac{5}{100} + \frac{2}{100} = -\frac{3}{100} \] So, \[ -\frac{3}{100} = -4 \times 10^{-3} t_1 \] --- ## 5. Solve for \( t_1 \) \[ \frac{3}{100} = 4 \times 10^{-3} t_1 \] \[ t_1 = \frac{3}{100} \div 4 \times 10^{-3} = \frac{3}{100} \times \frac{1}{4 \times 10^{-3}} \] Calculate: First, \( \frac{1}{4 \times 10^{-3}} = \frac{1}{0.004} = 250 \). So, \[ t_1 = \frac{3}{100} \times 250 = 3 \times 2.5 = 7.5 \] --- ## **Final Answer** \[ \boxed{t = 7.5 \text{ seconds}} \] So, it takes **7.5 seconds** for the airplane to reduce its speed from 50 m/sec to 20 m/sec under the given acceleration.