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Questions 1 to 7 from RACE #17 Physics
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( \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \right) dx = \frac{3}{4} x^{4/3} - \frac{3}{2} x^{2/3} + C \\ 6. \int 3x dx = \frac{3x^2}{2} + C \\ 7. \int 5x^2 dx = \frac{5x^3}{3} + C \\ \text{2. Distance travelled} = 39 \text{ m (Option B)} \\ \text{3. } t = 2 \text{ s (Option B)} \\ \text{4. Particle stops at } t=1 \text{ s and continues without reversing (Option B)} \\ \text{5. Maximum time from A to B} = \frac{(\sqrt{3}+1)}{4} T \text{ (Option A)} \\ \text{6. Distance } s = ut + \frac{kt^3}{6} \text{ (Option B)} \\ \text{7. Total distance } = \frac{m V_0}{r} \text{ (Option A)} \end{cases}
1. Integrals as above. 2. Distance = 39 m. 3. t = 2 s. 4. Stops at t=1 s and continues without reversing. 5. Max time from A to B = ((√3+1)/4) T. 6. s = ut + (kt^3)/6. 7. Distance = (m V_0)/r.