Abluefin tuna can reach speeds of up to 22 m/s. suppose one tuna falls behind the rest of the school by a distance of 55 meters. how quickly can the tuna catch up a)1.75 seconds b)2.5 seconds c)3.45 seconds d)4.3 seconds

We know that velocity is distance traveled over a period of time or v=d/t. Now what the problem is asking is how quickly or how long it will take the tuna to catch up the 55-meter gap if it can swim up to a velocity of 22m/s. 

We can solve this using the velocity formula. Fill up what you know and solve for what you do not know. We know that v = 22m/s and the distance that needs to be traveled is 55m. 

v = d/t
22m/s = 55m/t

We need to first isolate time on one side of the equation. You can do so by first transposing it to the other side of the equation. When you transpose, you need to remember that whatever operation you will use on one side, you do the opposite on the other side. Since time uses division on one side, we use multiplication on the other side.  

(22m/s)(t) = 55m    Now that we got it to the other side, we need to isolate it                                 by transposing the velocity, following the same rule                                         of using the opposite operation. 
t = (55m)/(22m/s)   The next thing to do is to do the operation.
t = 2.5s                    

It will take the tuna 2.5 s to catch up. The answer is B. 

Why the rest of the question?