CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
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(sin370.6) 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! We‘ve got your back. Try BYJU‘S free classes today!
B
4min
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3min
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 4min
VSR= Velocity of swimmer w.r.t river=5km/hr
VR= Velocity of river w.r.t ground
VS= Velocity of swimmer w.r.t ground
VS=VSR+VR
VS=(VRVSRsin370)ˆi+VSRcos370ˆj
Displacement along x-axis=
d=(VRVSRsin370)(0.05)
=[45(35)](0.05)
=0.05km
=50m
Total time=50×103km3km/hr+0.05hr
=0.053+0.05=0.203=0.0666hr
=4min

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relative Motion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon