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Step-by-Step Solution
Step 1: Identify the known information
• A bullet has an initial velocity $v_i$.
• After traveling 4 cm ($0.04$ m), it loses one-third of its velocity, meaning its new velocity is $\tfrac{2}{3}v_i$.
• It then travels an additional distance of $D \times 10^{-3}$ m before coming to rest.
• We assume uniform deceleration throughout its motion.
Step 2: Write the kinematic equation for the first 4 cm
We use the kinematic relation $v^2 = u^2 + 2as$. Here,
• $u = v_i$,
• $v = \tfrac{2}{3}v_i$,
• $s = 0.04$ m,
• $a$ is the deceleration (a negative quantity).
So,
$$
\left(\tfrac{2}{3}v_i\right)^2 = v_i^2 + 2 \, a \, \bigl(0.04\bigr).
$$
Step 3: Solve for the deceleration $a$
Calculate the left-hand side:
$$
\tfrac{4}{9}v_i^2 = v_i^2 + 2 \, a \, (0.04).
$$
Rearrange to isolate $a$:
$$
2 \, a \, (0.04) = \tfrac{4}{9}v_i^2 - v_i^2
= v_i^2 \Bigl(\tfrac{4}{9} - 1\Bigr)
= -\tfrac{5}{9}v_i^2.
$$
Thus,
$$
a = \frac{-\tfrac{5}{9}v_i^2}{2 \times 0.04}
= -\tfrac{5}{9} \,\frac{v_i^2}{0.08}.
$$
Step 4: Apply the kinematic equation to the remaining distance
Next, the bullet goes from $\tfrac{2}{3}v_i$ to $0$ over a distance $D \times 10^{-3}$ m. Again using $v^2 = u^2 + 2as$:
$$
0 = \bigl(\tfrac{2}{3}v_i\bigr)^2 + 2 \, a \, \bigl(D \times 10^{-3}\bigr).
$$
So,
$$
\bigl(\tfrac{2}{3}v_i\bigr)^2 + 2 \, a \, \bigl(D \times 10^{-3}\bigr) = 0.
$$
Step 5: Substitute $a$ and find $D$
We already have $a = -\tfrac{5}{9}\,\frac{v_i^2}{0.08}$. Substitute into the equation from Step 4:
$$
\tfrac{4}{9}v_i^2 + 2 \Bigl(-\tfrac{5}{9}\,\tfrac{v_i^2}{0.08}\Bigr)\, \bigl(D \times 10^{-3}\bigr) = 0.
$$
Rearrange:
$$
2 \Bigl(-\tfrac{5}{9}\,\tfrac{v_i^2}{0.08}\Bigr)\bigl(D \times 10^{-3}\bigr) = -\tfrac{4}{9}v_i^2.
$$
Cancelling $v_i^2$ and the negative signs, and noting $\frac{1}{0.08} = 12.5$, gives:
$$
\tfrac{10}{9}\,\tfrac{1}{0.08} \,\bigl(D \times 10^{-3}\bigr) = \tfrac{4}{9}
\quad \Rightarrow \quad
\tfrac{10}{9}\times 12.5 \,\bigl(D \times 10^{-3}\bigr) = \tfrac{4}{9}.
$$
So,
$$
\frac{125}{9}\,\bigl(D \times 10^{-3}\bigr) = \frac{4}{9}.
$$
Cancelling $\tfrac{1}{9}$ from both sides:
$$
125\,\bigl(D \times 10^{-3}\bigr) = 4.
$$
Thus,
$$
D \times 10^{-3} = \frac{4}{125} = 0.032\,\text{m},
$$
and therefore,
$$
D = 32.
$$
Step 6: Final answer
The value of $D$ is 32.
