Question

A boat covers a certain distance between two spots in a river taking  $ {t_1}\;{\text{hrs}} $  going downstream and  $ {t_2}\;{\text{hrs}} $  going upstream. What time will be taken by boat to cover the same distance in still water?

(A)  $ \dfrac{{{t_1} + {t_2}}}{2} $ 

(B)  $ 2\left( {{t_2} + {t_1}} \right) $ 

(C)  $ \dfrac{{2{t_1}{t_2}}}{{{t_1} + {t_2}}} $ 

(D)  $ \sqrt {{t_1}{t_2}}  $ 

Answer 1

Hint: As we know that if velocities are in the same direction then the resultant velocity will be the addition of the two and if the velocities are in the opposite direction then the resultant velocity will be subtraction of the two. If a boat is going upstream then its resultant velocity will be the difference of boat velocity and river steam velocity.

Complete step by step answer

In this question, the time taken by the boat to cover a certain distance between two spots in a river going downstream is  $ {t_1} $  and the time taken by the boat to cover a certain distance between two spots in a river going upstream is  $ {t_2} $ . We need to calculate the time taken by the boat to cover the same distance in the still water.

Let us assume, the velocity of the boat in the still water is  $ u $  the velocity of the river stream is  $ v $ , and the distance between the two spots is  $ d $ .

In downstream, the direction of the speed of the boat and the direction of the speed of the stream will be same, so the resultant velocity of the boat in the downstream is  $ \left( {u + v} \right) $  and it can be written as,

 $ {\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}} $ 

Now, substitute the value of the speed, distance, and the time in the above equation as,

 $  \Rightarrow \left( {u + v} \right) = \dfrac{d}{{{t_1}}}......\left( 1 \right) $ 

But, in upstream the direction of the speed of the boat is in opposite to the direction of the speed of the stream, so the resultant velocity of the boat in upstream is  $ \left( {u - v} \right) $  and it can be written as,

 $ {\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}} $ 

Now, substitute the value of the speed, distance, and the time in the above equation as,

 $  \Rightarrow \left( {u - v} \right) = \dfrac{d}{{{t_2}}}......\left( 2 \right) $ 

Now, add equation (1) and (2) to obtain the velocity of the boat in the still water as,

 $ \left( {u + v} \right) + \left( {u - v} \right) = \dfrac{d}{{{t_1}}} + \dfrac{d}{{{t_2}}} $ 

Now, we simplify the above equation as,

 $  \Rightarrow 2u = \dfrac{{d{t_2} + d{t_1}}}{{{t_1}{t_2}}} $ 

After simplification, the equation for the velocity of the boat in the still water will be,

 $  \Rightarrow u = \dfrac{{d\left( {{t_2} + {t_1}} \right)}}{{2{t_1}{t_2}}} $ 

Let us assume the time taken by the boat to cover the same distance in the still water be  $ t\;{\text{hrs}} $  and it is calculated as,

 $ {\text{Time}}\;{\text{taken}} = \dfrac{{{\text{Distance}}}}{{{\text{Velocity}}}} $ 

Now, write it is in variable form,

 $  \Rightarrow t = \dfrac{d}{u} $ 

Now, we substitute the expression of the velocity of the boat in the still water as,

 $ t = \dfrac{d}{{\dfrac{{d\left( {{t_2} + {t_1}} \right)}}{{2{t_1}{t_2}}}}} $ 

After simplification we get,

 $  \therefore t = \dfrac{{2{t_1}{t_2}}}{{{t_2} + {t_1}}} $ 

So, from the above calculation, the time taken by the boat to cover the same distance in the still water be  $ \dfrac{{2{t_1}{t_2}}}{{{t_2} + {t_1}}} $

Therefore, the correct option is (C).

Note: As we know that the velocity is the vector quantity and the resultant of the velocities will be calculated by using the vector addition, that is it depends on the direction of the velocities.