Question

A man rows upstream 16 km and downstream 27 km taking 5 hours each time. What is the velocity of current?

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Let the velocity of boat be "v",

and, the velocity of current be "u".

$\huge{\bold{\underline{Explanation:-}}}$

• Case-1

Downstream:-

relative velocity = (v + u)

(As the boat and current flowing in same direction so, their velocities will get added)

${ Distance = (Relative \: velocity) \times time \: taken}$

Substituting the values.

${ 27 = (v + u) \times 5}$

${ \to (v + u) = 5.4 -----(1)}$

(as 27/5 is 5.4)

• Case-2

Upstream:-

relative velocity = (v - u)

(As the boat and current flowing in opposite direction so, their velocities will get Subtracted)

${ Distance = (Relative \: velocity) \times time \: taken}$

Substituting the values.

${ 16 = (v - u) \times 5}$

${ \to (v - u) = 3.2 -----(2)}$

(as 16/5 is 3.2)

• Subtract (1) and (2)

${ \to 2u = 5.4 - 3.2}$

${ \to 2u = 2.2 }$

${ \to u = \frac{2.2}{2}}$

${ \to u = 1.1 km/hr}$

$\huge{\boxed{\boxed{ u = 1.1 \: km/hr}}}$

So,the velocity of current is 1.1 km/hr.

Answer 2

$\huge{\underline{\underline{\sf{Answer :}}}}$

Upstream

In upstream,the velocity of water current is in different direction as the velocity of boat. i.e.,velocities will be subtracted

• velocity would be v - u

Here,

• Distance Travelled,s = 16Km

• Time taken,t = 5h

Now,

$\sf{velocity = \frac{displacement}{time} } \\ \\ \rightarrow \: \sf{v + u = \frac{16}{5} } \\ \\ \rightarrow \: \sf{v - u = 3.2...........(1)}$

Downstream

While downstream,the direction of velocity of water current would be same to that of velocity of water. i.e.,they will be subtracted

• velocity would be v + u

Here,

• Distance Travelled,s' = 27Km

• Time taken,t = 5hr

Now,

$\sf{v - u = \frac{27}{5} } \\ \\ \rightarrow \: \sf{v + u = 5.4.............(2)}$

Adding equations (1) and (2),we get:

$\sf{2v = 8.6} \\ \\ \implies \: \boxed{ \sf{v = 4.3Kmh {}^{ - 1} }}$

Putting v in equation (1),we get:

$\sf{4.3 - u = 3.2 } \\ \\ \implies \: \boxed{ \sf{u = 1.1Kmh {}^{ - 1} }}$