1. **State the problem:** A bullet penetrates 0.04 m into a wall and loses 75% of its velocity. We need to find how far the bullet will penetrate the wall in total. 2. **Given:** - Initial velocity before penetration: $u$ - Velocity after penetrating 0.04 m: $v = 0.25u$ (since it loses 75%, it retains 25%) - Distance penetrated so far: $s_1 = 0.04$ m - Acceleration (deceleration) is constant, unknown 3. **Formula used:** The equation of motion $$v^2 = u^2 + 2as$$ where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, and $s$ is displacement. 4. **Step 1: Find acceleration $a$ during first 0.04 m penetration.** Using $v^2 = u^2 + 2as_1$: $$ (0.25u)^2 = u^2 + 2a(0.04) $$ $$ 0.0625u^2 = u^2 + 0.08a $$ Rearranged: $$ 0.08a = 0.0625u^2 - u^2 = -0.9375u^2 $$ $$ a = \frac{-0.9375u^2}{0.08} = -11.71875u^2 $$ 5. **Step 2: Find total penetration distance $S$ when velocity becomes zero.** At full stop, $v=0$, initial velocity $u$, acceleration $a$ as above, displacement $S$: $$ 0 = u^2 + 2aS $$ $$ 2aS = -u^2 $$ $$ S = \frac{-u^2}{2a} $$ Substitute $a$: $$ S = \frac{-u^2}{2(-11.71875u^2)} = \frac{1}{23.4375} \approx 0.0427 \text{ m} $$ 6. **Step 3: Interpret result.** Total penetration distance $S \approx 0.0427$ m. Since the bullet already penetrated 0.04 m, the remaining penetration is: $$ 0.0427 - 0.04 = 0.0027 \text{ m} $$ **Final answer:** The bullet will penetrate approximately $0.0427$ meters in total before stopping.