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

Answer:

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

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:

$t=\frac{d_{1} }{V_{BGup} } -\frac{d_{2} }{V_{BGdown} }\\t=\frac{1000m}{0.95m/s}-\frac{1000m}{2.05m/s}\\ t=564.83seconds=9.414minutes=0.1569hours$