Question

28. A sailor goes 8 km downstream in 40 minutes
and returns in 1 hour. Determine the speed of the
sailor in still water and the speed of the current.
​

Answer 1

Answer:

Let the speed of the sailor in still water be x km/hr and speed of the current be y km/hr.

Then, speed of sailor downstream =x+y km/hr and speed of sailor upstream =x−y km/hr.

Distance covered =8 km.

Also, Time= Distance÷speed

From the given information, we have,

x+y ÷ 8

=

60

40

and

x−y

8

=1

or,

x+y

8

=

60

40

=>

x+y

8

=

3

2

=>24=2(x+y)

=>x+y=12....(i)

Also,

x−y

8

=1

=>x=8+y....(ii)

Substituting (ii) in (i), we get,

x+y=12

=>8+y+y=12

=>2y=4

=>y=2

Substituting y=2 in equation (ii), we get,

x=8+y

=>x=8+2

=>x=10

Therefore, speed of sailor in still water =x=10 km/hr

and speed of the current =y=2 km/hr.

Answer 2

Answer:

speed of sailor = 10 kmph

speed of current = 2kmph

Step-by-step explanation:

speed of sailor = u

speed of current = v

downstream :

relative speed of sailor = u+v

t= 40 min = 2/3hr

u+v = dist/time = 8/(2/3) = 12 kmph ------(1)

upstream :

relative speed = u-v

t= 1 hr

u-v = 8/1= 8 kmph ----(2)

add (1) &(2)

2u = 20

u = 10 kmph i.e. speed of sailor

v= 2 kmph i.e speed of current