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JEE 2024 RACE #17 Physics Questions 1 to 7
Physics
\boxed{\begin{cases}
1. \int 3x^7 dx = \frac{3}{8} x^8 + C \\
2. \int 4\sqrt{x} dx = \frac{8}{3} x^{3/2} + C \\
3. \int \frac{dx}{\sqrt{x}} = 2 x^{1/2} + C \\
4. \int (x^3 - 5x^2 + 7x - 11) dx = \frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 11x + C \\
5. \int \left( x^{1/3} - x^{-1/3} \right) dx = \frac{3}{4} x^{4/3} - \frac{3}{2} x^{2/3} + C \\
6. \int 3x dx = \frac{3}{2} x^2 + C \\
7. \int 5x^2 dx = \frac{5}{3} x^3 + C \\
\text{2. Distance travelled} = 39 \text{ m} \\
\text{3. Time when } \vec{a} \perp \vec{v} = 2 \text{ s} \\
\text{4. Particle stops at } t=1 \text{ s and continues without reversing} \\
\text{5. Max time from A to B} = \frac{(\sqrt{3}+1)}{4} T \\
\text{6. Distance } s = ut + \frac{kt^3}{6} \\
\text{7. Total distance covered} = \frac{m V_0}{r}
\end{cases}
1. \( \frac{3}{8} x^8 + C, \frac{8}{3} x^{3/2} + C, 2 x^{1/2} + C, \frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 11x + C, \frac{3}{4} x^{4/3} - \frac{3}{2} x^{2/3} + C, \frac{3}{2} x^2 + C, \frac{5}{3} x^3 + C \); 2. 39 m; 3. 2 s; 4. Stops at 1 s, no reversal; 5. \( \frac{(\sqrt{3}+1)}{4} T \); 6. \( s = ut + \frac{kt^3}{6} \); 7. \( \frac{m V_0}{r} \).