In an experiment, a shearwater (a seabird) was taken from its nes...

Question

In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 km away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin at the nest and extend the +x–axis to the release point, what was the bird’s average velocity in m/s (a) for the return flight

Accepted Answer

welcome back everybody. We have a dog which I am no artist, but I sure will try. There we go, that'll work. And he runs 8.2 km towards a park and then he turns around and comes all the way back. Now the time for him to make the entire round trip trek is 25 minutes and we are tasked with finding what is the average velocity in meters per second. So we're gonna have to do some conversions here. Just as a reminder, the formula for average velocity is delta X over delta. Now it's important to note that we are finding the average velocity for the trip from the park to the cage. Have that little note there. Alright, so let's go ahead and start with these conversions. So we're gonna start with delta X, Delta X right now is equal to 8.2 km. Now There are m one kilometer units are gonna cancel out and we get m. However important note, we're going the other way. So this number is actually going to be negative 8200 m since we're going from the park back to the cage. Great. So now let's go ahead and convert our time here. So right now we have 25 minutes. Well, we know there are 60 seconds in one minute Units are gonna cancel out and we get seconds now that we have delta X and delta T in the correct units, we are ready to calculate the average velocity, which just one more time, Delta X divided by delta T. Equal to negative over 1500, which when you plug into your calculator, you get negative 5. m/s, which corresponds to our answer choice of B. Thank you guys so much for watching. Hope this video helped, and we will see you all in the next one.

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