1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A swimmer swims in still water at a speed =5km/hr. He enters a 200m wide river, having river flow speed = 4km/hr at a point A and proceeds to swim at an angle of 1270(sin37âˆ—âˆ’0.6) with the river flow direction. Another point B is located directly across A on the other side. The swimmer lands on the river bank at a point C. from which he walks the distance CB with a speed=3km/hr. The total time in which he reach from A to B is

A
5min
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
4min
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3min
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
None
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is B 4minVSR= Velocity of swimmer w.r.t river=5km/hrVR= Velocity of river w.r.t groundVS= Velocity of swimmer w.r.t groundVS=VSR+VRVS=(VR−VSRsin370)ˆi+VSRcos370ˆjDisplacement along x-axis=d=(VR−VSRsin370)(0.05)=[4−5(35)](0.05)=0.05km=50mTotal time=50×10−3km3km/hr+0.05hr=0.053+0.05=0.203=0.0666hr =4min

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Relative Motion
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program