Previous Questions

1. Integrals solved as per power rule. 2. Distance travelled = 39 m. 3. Time when acceleration ⟂ velocity = 2 s. 4. Particle stops at t=1 s and continues without reversing. 5. Max time from A to B = ((√3+1)/4) T. 6. Distance s = ut + (kt^3)/6. 7. Distance covered by boat = (mV₀)/r.

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. \( \frac{m V_0}{r} \).

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.

39 m

The ratio of velocities \( \frac{u_1}{u_2} = \frac{2}{\sqrt{3}} \) for the two bodies to collide.