Let the width of the river be \( d \).
Case 1: Shortest Path To cross the river along the shortest path (perpendicular to the river flow), the boat must counteract the river's flow. The effective velocity of the boat perpendicular to the river flow is: \[ V_{{eff}} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \, {ms}^{-1} \] The time taken to cross the river along the shortest path is: \[ t_1 = \frac{d}{12} \] Case 2: Shortest Time To cross the river in the shortest time, the boat should head directly across the river (perpendicular to the flow) without counteracting the flow. The effective velocity of the boat is simply its velocity in still water: \[ V_{{eff}} = 13 \, {ms}^{-1} \] The time taken to cross the river in the shortest time is: \[ t_2 = \frac{d}{13} \] Ratio of Times The ratio of the times taken is: \[ \frac{t_1}{t_2} = \frac{\frac{d}{12}}{\frac{d}{13}} = \frac{13}{12} \] Thus, the ratio is 13:12.
Final Answer: 13:12
A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is
If the roots of $\sqrt{\frac{1 - y}{y}} + \sqrt{\frac{y}{1 - y}} = \frac{5}{2}$ are $\alpha$ and $\beta$ ($\beta > \alpha$) and the equation $(\alpha + \beta)x^4 - 25\alpha \beta x^2 + (\gamma + \beta - \alpha) = 0$ has real roots, then a possible value of $y$ is: