A car drives down a road in such a way that its velocity ( in m/s) at time t (seconds) is v(t)=2t^1/2+5 Find the car's average velocity (in m/s) between t=2 and t=5.

Question

A car drives down a road in such a way that its velocity ( in m/s) at time t (seconds) is v(t)=2t^1/2+5

Find the car's average velocity (in m/s) between t=2 and t=5.

Could someone please assist with this question?

I asked this question but did not receive a correct answer as of yet. Any help and an explanation would be greatly appreciated. Thank you.

Answer 1

The average velocity is the total distance traveled divided by the total time transpired. To find the total distance traveled, given that you were given the velocity as a function of time, you can integrate the velocity function from t = 2 to t = 5 or:

To integrate, do the opposite of taking the derivative, in other words, what function, when I take the derivative, will be 2t^1/2 + 5. I'll assume you know how to do that. If not, you can plug the above integral into your TI-84 and get an answer.

The integral turns out to be (4/3)t^3/2 + 5t so x(t) = (4/3)t^3/2 + 5t and x(5) - x(2) = (4/3)(5)^3/2 + 5(5) - [(4/3)(2)^3/2 + 5(2)] = 26.136 meters

Now, the total travel was 26.136 meters and it was done in 3 seconds (5 - 2) so the average velocity is 26.136/3 = 8.712 m/s. I'm unsure if you need to worry about significant figures.