Question

water flows in a river with a velocity of 5km/h and a boat moves through the river with a speed of 7km/h in a direction perpendicular to the direction of water flow . what will be the resultant velocity of the river also calculate the angle made by the boat with the direction of water flow​

Answer 1

$\text{Upon drawing the given scenario we get the image attached}$

$\implies \text{The velocity of the river}(v_r) = 5 \: km/h = \dfrac{25}{18} \; m/s$

$\implies \text{The velocity of the boat}(v_b) = 7 \; km/h = \dfrac{35}{18} \; m/s$

$\implies \text{The resultant velocity of the boat}(\bar{v}) = v_r\hat{i} + v_b\hat{j}$

$\implies \boxed{\bar{v} = \dfrac{25}{18}\hat{i} + \dfrac{35}{18}\hat{j}}$

$\implies |\bar{v}| = \sqrt{(v_r)^{2} + (v_b)^{2} + 2v_rv_bcos\phi}$

$\implies \text{In this case }\phi=\dfrac{\pi}{2}$

$\implies cos\phi = 0$

$\implies |\bar{v}| = \sqrt{(v_r)^{2} +( v_b)^{2}}$

$\implies |\bar{v}| = \sqrt{\text{\huge{(}}\dfrac{25}{18}\text{\huge{)}}^{2} + \text{\huge{(}}\dfrac{35}{18}\text{\huge{)}}^{2}}$

$\implies |\bar{v}| = \sqrt{\dfrac{625}{324} + \dfrac{1225}{324}}$

$\implies |\bar{v}| = \sqrt{\dfrac{1850}{324}}$

$\implies \boxed{|\bar{v}| = \dfrac{5\sqrt{74}}{18} \; m/s = \sqrt{74} \; km/h}$

$\implies \text{The angle between }v\text{ and }v_b = \theta$

$\implies tan\theta = \dfrac{5}{7}$

$\implies \theta = tan^{-1}\text{\huge{(}}\dfrac{5}{7}\text{\huge{)}}$

$\implies \text{Angle made by }v \text{ and }v_r = \text{\huge{(}}\dfrac{\pi}{2} - \theta\text{\huge{)}} = \alpha$

$\implies \alpha =\dfrac{\pi}{2} - tan^{-1}\text{\huge{(}}\dfrac{5}{7}\text{\huge{)}}$

$\text{We know that }tan^{-1}x + cot^{-1}x = \dfrac{\pi}{2}$

$\implies \boxed{\alpha = cot^{-1}\text{\huge{(}}\dfrac{5}{7}\text{\huge{)}}}$

$\text{We know that }cot^{-1}x = tan^{-1}\text{\huge{(}}\dfrac{1}{x}\text{\huge{)}}$

$\implies \boxed{\alpha = tan^{-1}\text{\huge{(}}\dfrac{7}{5}\text{\huge{)}}}$