A bullet is fired horizontally over level ground with a muzzle velocity of \( 1500 \mathrm{~m} / \mathrm{s} \) and the retardation (negative acceleration) it experiences ofter \( t \) seconds is given by the formula \[ a=-\frac{9000}{(t+5)^{2}} \] Derive formulae for the velocity and displacement of the bullet and then calculate (to 4 significant digits) the velocity and displacement of the bullet after \( 1,000 \mathrm{~s} \).

To begin with, let's derive the formulae for the velocity and displacement of the bullet. 1. **Velocity Formula**: The acceleration \( a(t) \) is given as: \[ a = -\frac{9000}{(t+5)^{2}} \] Since acceleration is the derivative of velocity with respect to time, we can express the velocity \( v(t) \) as: \[ v(t) = v_0 + \int a(t) \, dt \] Substituting \( v_0 = 1500 \, \mathrm{m/s} \): \[ v(t) = 1500 + \int -\frac{9000}{(t+5)^{2}} \, dt \] Performing the integration: \[ \int -\frac{9000}{(t+5)^{2}} \, dt = \frac{9000}{t+5} + C \] where \( C \) is a constant. To find \( C \), we note that at \( t = 0 \), \( v(0) = 1500 \): \[ v(0) = 1500 + \frac{9000}{5} + C \] Hence, \( C = 1500 - 1800 = -300 \). Thus, the velocity formula is: \[ v(t) = 1500 - \frac{9000}{t+5} \] 2. **Displacement Formula**: The displacement \( s(t) \) is found by integrating the velocity: \[ s(t) = \int v(t) \, dt = \int \left( 1500 - \frac{9000}{t+5} \right) \, dt \] \[ s(t) = 1500t - 9000 \ln(t+5) + C' \] To find the constant \( C' \), let's assume \( s(0) = 0 \): \[ 0 = 1500(0) - 9000 \ln(5) + C' \] Thus, \( C' = 9000 \ln(5) \) and the displacement formula becomes: \[ s(t) = 1500t - 9000 \ln(t+5) + 9000 \ln(5) \] 3. **Calculating Velocity and Displacement at \( t = 1000 \, \mathrm{s} \)**: - **Velocity \( v(1000) \)**: \[ v(1000) = 1500 - \frac{9000}{1000+5} = 1500 - \frac{9000}{1005} \] \[ v(1000) \approx 1500 - 8.95 \approx 1491.05 \, \mathrm{m/s} \] - **Displacement \( s(1000) \)**: \[ s(1000) = 1500(1000) - 9000 \ln(1005) + 9000 \ln(5) \] First, calculate \( \ln(1005) \) and \(\ln(5)\): \[ \ln(1005) \approx 6.912 \quad \text{and} \quad \ln(5) \approx 1.609 \] Thus, \[ s(1000) = 1500000 - 9000 \times 6.912 + 9000 \times 1.609 \] \[ s(1000) = 1500000 - 62108 + 14481 \approx 1500000 - 47627 \approx 1452373 \, \mathrm{m} \] In conclusion, after \( 1000 \, \mathrm{s} \): - The velocity is approximately \( 1491.1 \, \mathrm{m/s} \) (to 4 significant digits). - The displacement is approximately \( 1.452 \times 10^6 \, \mathrm{m} \) (or about \( 1452000 \, \mathrm{m} \) to 4 significant digits).