The distance (in feet) of an object from a point is given by s(t)=t2, where time t is in seconds.
(a) what is the average velocity of the object between t=9 and t=11?
(b) by using smaller and smaller intervals around 9, estimate the instantaneous velocity at time t=9⁢ . round your answer to the nearest integer.

Step-by-step explanation:

If we plotted this on a position-v-time graph in physics, the slope of the line between the times 9 and 11 seconds is the average velocity of the object.  If we do not have the capabilities of calculus to figure out the derivative (which will give us the instantaneous velocity equation of the object) we have to use the position function and use the slope formula to find the velocity that way.  If s(t) = t², then s(9) = 81 and the coordinate for this is (9, 81).  If we plug in 11 for the time, s(11) = 121 and the coordinate for that is (11, 121).  Applying the slope formula:

So the average velocity is 20 ft per second

The idea of instantaneous velocity is to get really really super close to 9 seconds and then find the instantaneous velocity using the same formula.  However, because instantaneous velocity is found at a SINGLE point in time, we cannot use the formula because we don't have 2 points to plug in.  That's where calculus comes in handy.  To get close to the instantaneous velocity, I used times 8.9 seconds and 9.1 seconds.

s(8.9) = 79.21 and

s(9.1) = 82.81

Now if we plug into the slope formula:

So the average velocity at almost exactly 9 seconds is 18 feet per second.  When you learn derivatives in calculus, you will find that the instantaneous velocity is, coincidentally, 18 feet per second.