Given:
The bullet loses one-third of its velocity after traveling \(s = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m}\), with final velocity \(v = \frac{2u}{3}\).
Using the kinematic equation:
\(v^2 - u^2 = 2a \cdot s\)
Substituting the values:
\(\left(\frac{2u}{3}\right)^2 - u^2 = 2a \cdot (4 \times 10^{-2})\)
Simplifying:
\(\frac{4u^2}{9} - u^2 = 2a \cdot (4 \times 10^{-2})\)
\(\frac{4u^2}{9} - \frac{9u^2}{9} = 2a \cdot (4 \times 10^{-2})\)
\(-\frac{5u^2}{9} = 2a \cdot (4 \times 10^{-2})\)
\(a = \frac{-5u^2}{72 \times 10^{-2}}\)
Now, for the bullet to come to rest:
\(v^2 - u^2 = 2a \cdot D\)
Substitute \(v = 0\), \(a = \frac{-5u^2}{72 \times 10^{-2}}\), and solve for \(D\):
\(0 - u^2 = 2 \cdot \left(\frac{-5u^2}{72 \times 10^{-2}}\right) \cdot D\)
\(u^2 = \frac{10u^2}{72 \times 10^{-2}} \cdot D\)
\(D = \frac{72 \times 10^{-2}}{10}\)
\(D = 32 \times 10^{-3} \, \text{m} = 32 \, \text{mm}.\)
The Correct answer is: 32mm
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: