Question

In a 200.0-m relay race (each leg of the race is 50.0 m long), one swimmer has a 0.450 second lead

Answer 1

The minimum  speed that the second swimmer must have in order to catch up with the first swimmer by the end of the  pool is; 4.04 m/s

We are given the following facts;

• Both swimmers swim the same distance of 50 m for a leg of the relay race.

• Swimmer 1 swims at constant speed of 3.9 m/s.

• Swimmer 1 has a 0.450 second head start.

Formula for time taken is;

time = distance/speed

Time taken by swimmer 1 for one leg;

t₁ = 50/3.9

t₁ = 12.82 s

Since swimmer 1 has a head start of 0.45 s, then time that swimmer 2 must use in order to catch up is;

t₂ = 12.82 -0.45

t₂ = 12.37 s

Thus, speed of swimmer 2;

v = 50/12.37

v = 4.04 m/s

Read more about distance, speed and time at; brainly.com/question/24571540