A swimmer swims downstream from point A to point B in 4 hours. It covers the same distance upstream in 5 hours. If the speed of the stream is 2 km/h, what is the distance between A and B:
80 km
This problem involves understanding how the speed of the stream affects the speed of a swimmer moving with or against the current. We need to use the concepts of relative speed to find the distance between two points.
Let's define the variables:
We know that Distance $=$ Speed $\times$ Time.
Since the distance $d$ is the same in both cases, we can set the two expressions for $d$ equal to each other:
$(s + 2) \times 4 = (s - 2) \times 5$
Now, we solve this equation to find the value of $s$, the speed of the swimmer in still water:
$4s + 8 = 5s - 10$
To solve for $s$, we can rearrange the terms:
$8 + 10 = 5s - 4s$
$18 = s$
So, the speed of the swimmer in still water is 18 km/h.
Now that we have the speed of the swimmer in still water ($s = 18$ km/h), we can use either the downstream or upstream equation for distance to find $d$.
Using the downstream equation:
$d = (s + 2) \times 4$
$d = (18 + 2) \times 4$
$d = 20 \times 4$
$d = 80$ km
Using the upstream equation (just to double-check):
$d = (s - 2) \times 5$
$d = (18 - 2) \times 5$
$d = 16 \times 5$
$d = 80$ km
Both calculations give the same distance, 80 km.
| Concept | Formula | Values |
|---|---|---|
| Downstream Speed | Speed of swimmer + Speed of stream | $s + 2$ km/h |
| Upstream Speed | Speed of swimmer - Speed of stream | $s - 2$ km/h |
| Downstream Distance | Downstream Speed $\times$ Downstream Time | $(s + 2) \times 4$ |
| Upstream Distance | Upstream Speed $\times$ Upstream Time | $(s - 2) \times 5$ |
| Equating Distances | $(s + 2) \times 4 = (s - 2) \times 5$ | $s = 18$ km/h |
| Final Distance | $(18 + 2) \times 4$ or $(18 - 2) \times 5$ | $80$ km |
The distance between A and B is 80 km.
| Term | Definition | Formula |
|---|---|---|
| Speed of Swimmer in Still Water ($s$) | The speed at which the swimmer moves without the influence of the current. | - |
| Speed of Stream ($v$) | The speed of the flowing water. | - |
| Downstream Speed | The effective speed when moving with the current. | $s + v$ |
| Upstream Speed | The effective speed when moving against the current. | $s - v$ |
| Relationship | If speeds are $S_D$ (downstream) and $S_U$ (upstream), then $s = \frac{S_D + S_U}{2}$ and $v = \frac{S_D - S_U}{2}$. | - |
Problems involving relative speed in water (like swimmers or boats in streams) are common applications of the basic speed, time, and distance formula ($D = S \times T$).
Key points to remember:
These problems often require setting up and solving linear equations, as demonstrated in the solution above.
The speed of a ship in still water is 5 km/hr and the speed of the stream is 2 km/hr. Rohan rows to place at a distance of 21 km and comes back to the starting point. The total time taken by him is:
The speed of a boat in still water is 9 km/hr and the speed of stream is 3 km/hr. The difference between the upstream speed and downstream speed will be:
A boat can go 10 km upstream and 11 km downstream in a total time of 52 minutes, If the speed of the stream is 5 km/h, then what is the speed (in km/h) of the boat when going downstream?
The upstream speed of the boat is 40 km/hr and the speed of the boat in still water is 55 km/hr. What is the downstream speed of the boat?
A. 75 km/hr
B. 70 km/hr
C. 60 km/hr
D. 65 km/hrA boat moving upstream takes 8 hours 48 minutes to cover a distance while it takes 4 hours to return to the starting point, downstream. What is the ratio of the speed of boat in still water to that of water current?