A river of width 200 text m$200 \text{ m}$ is flowing from west to east with a speed of 18 text km/h$18 \text{ km/h}$. A boat, moving with speed of 36 text km/h$36 \text{ km/h}$ in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ________ and ________ respectively.

### Related Formula t_textmin = fracdv_br$$t_{\text{min}} = \frac{d}{v_{br}}$$ textDrift (displacement along bank) = v_r times t_texttotal$$\text{Drift (displacement along bank)} = v_r \times t_{\text{total}}$$ ### Core Logic First, convert speeds to m/s. River speed v_R = 18 times frac518 = 5 text m/s$v_R = 18 \times \frac{5}{18} = 5 \text{ m/s}$. Boat speed in still water v_BR = 36 times frac518 = 10 text m/s$v_{BR} = 36 \times \frac{5}{18} = 10 \text{ m/s}$. River width d = 200 text m$d = 200 \text{ m}$.

For minimum time across the river, the boat must head exactly perpendicular to the river flow. ### Step 1: Calculating Minimum Time Time taken for one crossing (bank to bank): t_textone-way = fracdv_BR = frac20010 = 20 text s$$t_{\text{one-way}} = \frac{d}{v_{BR}} = \frac{200}{10} = 20 \text{ s}$$ Since it is a round trip (bank to bank and back), total minimum time is: t_texttotal = 2 times 20 = 40 text s$$t_{\text{total}} = 2 \times 20 = 40 \text{ s}$$ ### Step 2: Calculating Displacement along Bank Displacement along the river bank (drift) happens exclusively due to the river's flow during this total time: textDisplacement = v_R times t_texttotal = 5 times 40 = 200 text m$$\text{Displacement} = v_R \times t_{\text{total}} = 5 \times 40 = 200 \text{ m}$$ ### Pattern Recognition Minimum time always ignores drift and maxes out the perpendicular velocity vector. Drift simply becomes V_textriver times T_texttotal$V_{\text{river}} \times T_{\text{total}}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics