Question

A boat covers the distance between two points in a river in 6 hours downstream and 8 hrs upstream.
A floating body in the lake crosses these two pionts in
[ ]
a) 8 hrs
b) 16 hrs
c) 18 hrs
d) 2 days​

Answer 1

Explanation:

Let velocity of water be u and velocity of boat in still water be v

Let distance travelled be d

We need to find d/v

t1= d/(v+u)

d= t1 v + t1 u (1)

t2= d/(v-u)

d = t2 v- t2 u (2)

From 1&2

v t1 + u t1= v t2- u t2

u(t1+ t2) = v(t2- t1)

u = v(t2-t1) / (t1+t2)

Substituting value of u in (1)

d = t1v + t1 [v(t2-t1) /t1+t2)] t1

d(t1+t2) = t1v(t1+t2) + vt1 (t2-t1)

d(t1+t2) = t1vt1 + t2t1v + vt1t2 - t1t1v

d(t1+t2) =2vt1t2

d/v = 2t1t2/( t1+t2)

Answer 2

Answer:

6.86hr

Explanation:

Let the velocity of a boat in still water is u and velocity of the river is v and distance is d

down the stream speed will add total velocity is u+v it takes time t1

in upstream speed will be subtracted total velocity is u-v it take time t2

u+v= d/t1 ...(I)

u−v= d/t2 ...(II)

2u = d/t1 + d/t2

u = d(t1+t2) / (2*t1*t2)

time take by the boat in still water is

t = d/u = (2*t1*t2) / (t1+t2)

now,

we have,

t1 = 6 hr

t2 = 8 hr

t = (2 * 6 * 8) / (6+8)

t = 48/7 hr = 6.86 hr