Average Velocity

Question

A particle moves on a line away from its initial position so that after t hours it is s = 4t^2 + t miles from its initial position. Find the average velocity of the particle over the interval [1, 4]. Include units in your answer

Answer 1

Take derivative of s getting v = 8t + 1, so v at t = a = 9 and v at t = 4 is 33. If you plot this on a graph you will get a straight line with slope of 8. Graphically the area under this curve between 1 and 4 is the distance covered. Set that area equal to v ( 4 - 1) = 3v and solve for v and you will get 21 miles per hour.

Answer 2

The velocity, v(t) , is ds/dt = 8t+1. At time 1, v1 = 9. At time 4, v4 = 33. So within this t interval [1,4], v starts at 9 and ends at 33 miles per hour. So the average velocity = (33 + 9)/2 = 21 miles per hour.

Another way to look at this problem is to see s(4) which is 68 and s(1) which is 5. (68-5)/(4-1) also = 21miles per hour.