Question

In the 2016 Olympics in Rio, after the 50 m freestyle competition, a problem with the pool was found. In lane 1there was a gentle 1.2 cm/s current flowing in the direction that the swimmers were going, while in lane 8there was a current of the same speed but directed opposite to the swimmers' direction. Suppose a swimmer could swim the 50.0 m in 25.0 s in the absence of any current. How would the time it took the swimmer to swim 50.0 m change in lane 1?
Enter negative value if the swimmer would be faster and positive value if the swimmer would be slower. Part B
How would the time it took the swimmer to swim 50.0 m change in lane 8?
Enter negative value if the swimmer would be faster and positive value if the swimmer would be slower.

Answer 1

A) -0.15 s

First of all, we need to velocity of the swimmer in absence of current. This is given by:

$v_0 = \frac{d}{t}$

where

d = 50.0 m

t = 25.0 s

Substituting,

$v_0 = \frac{50.0}{25.0}=2.0 m/s$

In lane 1, the velocity of the current is

$v_c = +1.2 cm/s = +0.012 m/s$

where the + sign means it is in the same direction as the swimmer. Therefore, the net velocity of the swimmer in lane 1 will be

$v=v_0+v_c = 2.0 + 0.012 = = 2.012 m/s$

And so, the time the swimmer will take to cover the 50.0 m will be:

$t=\frac{d}{v}=\frac{50.0}{2.012}=24.85 s$

So, the time would change by

$\Delta t = 24.85 - 25.0 = -0.15 s$

which means that the swimmer will be 0.15 s faster.

B) +0.15 s

To solve this part, we just need to consider that the current goes in the opposite direction, so its velocity actually is:

$v_c = -12 cm/s = -0.012 m/s$

where the negative sign indicates the opposite direction.

So, the net velocity of the swimmer in lane 8 is

$v=v_0+v_c = 2.0 - 0.012 = = 1.988 m/s$

And so, the time the swimmer will take to cover the 50.0 m will be:

$t=\frac{d}{v}=\frac{50.0}{1.988}=25.15 s$

So, the time would change by

$\Delta t = 25.15 - 25.0 = +0.15 s$

which means that the swimmer will be 0.15 s slower.