1. **Problem statement:** A bullet moving at an initial velocity of $600\ \text{m/s}$ penetrates a target to a depth of $5\ \text{cm}$ and slows down to $200\ \text{m/s}$. Assuming uniform resistance, we want to find how far the bullet will travel before coming to rest. 2. **Key idea:** The bullet experiences uniform deceleration due to resistance. We can use the kinematic equation relating velocities, acceleration, and displacement: $$v^2 = u^2 + 2as$$ where $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration (negative for deceleration), and $s$ is displacement. 3. **Step 1: Find acceleration during the first penetration.** Given: - Initial velocity $u_1 = 600\ \text{m/s}$ - Final velocity $v_1 = 200\ \text{m/s}$ - Displacement $s_1 = 5\ \text{cm} = 0.05\ \text{m}$ Using the formula: $$v_1^2 = u_1^2 + 2 a s_1$$ Rearranged to solve for $a$: $$a = \frac{v_1^2 - u_1^2}{2 s_1}$$ Calculate: $$a = \frac{200^2 - 600^2}{2 \times 0.05} = \frac{40000 - 360000}{0.1} = \frac{-320000}{0.1} = -3,200,000\ \text{m/s}^2$$ The acceleration (deceleration) is $-3,200,000\ \text{m/s}^2$. 4. **Step 2: Find the additional distance $s_2$ the bullet travels from $200\ \text{m/s}$ to rest ($0\ \text{m/s}$) under the same acceleration.** Given: - Initial velocity $u_2 = 200\ \text{m/s}$ - Final velocity $v_2 = 0\ \text{m/s}$ - Acceleration $a = -3,200,000\ \text{m/s}^2$ Using the formula: $$v_2^2 = u_2^2 + 2 a s_2$$ Rearranged: $$s_2 = \frac{v_2^2 - u_2^2}{2 a} = \frac{0 - 200^2}{2 \times (-3,200,000)} = \frac{-40000}{-6,400,000} = 0.00625\ \text{m}$$ 5. **Step 3: Total distance traveled before coming to rest** is the sum of the initial penetration and the additional distance: $$s_{total} = s_1 + s_2 = 0.05 + 0.00625 = 0.05625\ \text{m} = 5.625\ \text{cm}$$ **Final answer:** The bullet will travel approximately $5.625\ \text{cm}$ before coming to rest under uniform resistance.