A bullet travelling at a speed of 350 m/s enters a concrete wall and penetrates a distance of 5.0 cm before coming to rest. Determine the deceleration in m/s ^2 .

Step-by-step solution

Using the third equation of motion, v^2 = u^2 + 2as Given u = 350 m/s , v = 0 m/s , and s = 5.0 cm = 0.05 m Substituting the values: 0 = (350)^2 + 2a(0.05) 0 = 122500 + 0.1a a = -1225000 m/s ^2 The deceleration is the magnitude of negative acceleration, which is 1225000 m/s ^2 .