If the position of an ant traveling along a horizontal path at time t is 3t^2+1, what is the ant's average velocity from t=1 to t=6?

For this case, the first thing you should do is get the equation for speed. To do this, you derive the equation of the position with respect to time.

We have then:

r (t) = 3t ^ 2 + 1

r ' (t) = v (t) = 6t

We observe that the equation of velocity is a line.

Since it is a line, the average rate of change for any interval is given by the slope.

We have then that the average exchange rate is:

v = 6 m / s

Answer:

the ant's average velocity from t = 1 to t = 6 is:

v = 6 m / s