Question

A river has a steady speed of 0.550 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point. If the student can swim at a speed of 1.50 m/s in still water, how long does the trip take (swimming up and down in the current)?If the water were still, by how much would the trip be longer or shorter?

Answer 1

`t=564.83seconds=9.414minutes=0.1569hours`

Step-by-step explanation:

Here the steady speed of the river water relative to ground is

`V_(wg)=0.55m/s`

Speed of the boy relative to still water is

`V_(BW)=1.50m/s`

Here we are considering +ve speed along along +ve x-axis and -ve speed along -ve x-axis

Therefore the speed of boy relative to the ground upstream is:

`V_(BGup)=V_(BW)-V_(WG)\\V_(BGup)=1.50m/s-0.550m/s\\V_(BGup)=0.95m/s`

And the speed of boy relative to the ground downstream is:

`V_(BGdown)=V_(BW)+V_(WG)\\V_(BGdown)=1.50m/s+0.550m/s\\V_(BGdown)=2.05m/s`

The distance covered in upstream trip d₁=1.0km=1000m and the distance covered in downstream is d₂=1.0km=1000m

Since time taken by a person is to cover distance d with speed v given by:

`t=d/v`

The total time taken for one trip is:

`t=(d_(1) )/(V_(BGup) ) -(d_(2) )/(V_(BGdown) )\\t=(1000m)/(0.95m/s)-(1000m)/(2.05m/s)\\ t=564.83seconds=9.414minutes=0.1569hours\\`