Question

The difference between downstream speed and upstream speed is 3 km/hr and the total time taken during upstream and downstream is 3 hours. What is the downstream speed, if the downstream and upstream distance are 3 km each?

Answer 1

The difference between downstream speed and upstream speed is 3 km/hr and the total time taken during upstream and downstream is 3 hours. What is the downstream speed, if the downstream and upstream distance are 3 km each?

Let Say Speed of Stream = x km/hr

and speed of Rowing = y km/hr

upstream spedd = y - x m/he

Down stream speed = y + x km/Hr

Difference between downstream & upstram speed = 3 km/hr

y + x - (y -x) = 3

=> 2x = 3

x = 3/2 km/Hr

Speed of stream = 3/2 Km/hr

Downstream speed = y +3/2 km/Hr

Upstream speed = y - 3/2 km/hr

Total Time taken = 3/(y +3/2)  + 3/(y-3/2)  = 3

2/(2y +3)  + 2/(2y -3)  = 1

=> 4y - 6 + 4y + 6 = 4y² - 9

=> 4y² - 8y - 9 = 0

=> y = (8 + √64 + 144)/8

=> y = (8 + √208)/8

=> y = (8 + 14.42)/8

=> y = 22.42/8

=> y = 2.8 km/hr

Downstream speed = y +3/2 km/Hr

= 2.8 + 1.5

= 4.3 km/Hr

Answer 2

Given:

Difference between downstream speed and upstream speed $\[ = 3{\text{ }}km/hr\]$

Total time taken during upstream and downstream $\[ = 3{\text{ }}hours\]$

Downstream distance $\[ = 3km\]$

Upstream distance $\[ = 3{\text{ }}km\]$

To Find: Downstream speed

Solution:

Consider that the upstream speed is ${v_u}$ and the downstream speed is ${v_d}$.

Then,

$\begin{gathered} {v_d} - {v_u} = 3{{km} \mathord{\left/ {\vphantom {{km} h}} \right. \kern-\nulldelimiterspace} h} \hfill \\ {v_u} = {v_d} - 3km/h \hfill \\ \end{gathered}$

Suppose that the time taken in upstream is $t$ hours, then, the time taken in downstream will be $3-t$ hours.

Therefore,

$\[\begin{gathered} {v_u} = \frac{{3km}}{t} \hfill \\ \Rightarrow t = \frac{{3km}}{{{v_u}}} \hfill \\ \end{gathered} \]$

And,

$\begin{gathered} {v_d} = \frac{{3km}}{{3 - t}} \hfill \\ \Rightarrow {v_d} = \frac{{3km}}{{3 - \frac{{3km}}{{{v_u}}}}} \hfill \\ \Rightarrow {v_d} = \frac{{{v_u}}}{{{v_u} - 1}} \hfill \\ \end{gathered}$

Substituting the value of $v_{u}$ in the above equation, we get,

$\[\begin{gathered} {v_d} = \frac{{{v_d} - 3km/h}}{{{v_d} - 3km/h - 1}} \hfill \\ {v_d} = \frac{{{v_d} - 3}}{{{v_d} - 4}} \hfill \\ v_d^2 - 4{v_d} = {v_d} - 3 \hfill \\ v_d^2 - 5{v_d} + 3 = 0 \hfill \\ \end{gathered} \]$

Further solving, we have,

$\[{v_d} = \frac{{5 \pm \sqrt {{{\left( { - 5} \right)}^2} - 4 \times 1 \times 3} }}{{2 \times 1}}\]$

$\[{v_d} = \frac{{5 \pm \sqrt {13} }}{2}\]$

On taking the positive sign

$\[{v_d} = \frac{{5 + \sqrt {13} }}{2}\]$

$\[{v_d} = 4.3km/h\]$

Thus,

$\begin{gathered} {v_u} = 4.3km/h - 3km/h \hfill \\ {v_u} = 1.3km/h \hfill \\ \end{gathered}$

Similarly, on taking the negative sign,

$\[{v_d} = \frac{{5 - \sqrt {13} }}{2}\]$

$\[{v_d} = 0.7km/h\]$

Then,

$\begin{gathered} {v_u} = 0.7km/h - 3km/h \hfill \\ {v_u} = - 2.3km/h \hfill \\ \end{gathered}$

The negative value of upstream speed is not relevant, therefore, the possible downstream speed is $4.3\,km/h$.

Hence, the downstream speed is $4.3\,km/h$.

#SPJ2