If a rock is dropped from a height of 113 ft, its position t seconds after it is dropped until it hits the ground is given by the function s(t) = -16t^2 + 113.

Question

1. How long does it take the rock to hit the ground?
2. Find the average velocity of the rock from when it is released until when it hits the ground.
3. What time after the rock is thrown will its instantaneous velocity be equal to its average velocity? (Apply the Mean Value Theorem) Round answers to 3 decimal places

Answer 1

1. Let's consider the meaning of the s(t) function. This describes the height of the rock with respect to time. So, if we want to know when it hits the ground, we would want to know when the height = 0. So, just set your height function s(t) = 0. Find the time that will make it true!

0 = -16t² + 113

t² = -113/-16

t = 2.66 approx

2. The average velocity means the total distance traveled by the rock divided by the total time t. We know that it took 2.66 seconds to reach the ground, and we also know that it traveled -113 feet as it was dropped Thus

V_Average = -113/2.66 = -42.48 ft/sec

3. The question asks to find when the instantaneous velocity will match the average velocity (we just found to be -113/2.66). To find instantaneous velocity, we can find the derivative of the position function s(t).

s'(t) = -32t

The mean value theorem states that

f'(c) = [f(b)-f(a)] / (b-a)

where b and a are the time intervals in which the rock travels, so a = 0 to b = 2.66

So all we have to do is set treat our s'(t) like f'(c).

-32t = [f(b)-f(a)] / (b-a)

And we know that f(2.66) - f(0) = -113 and b-a = 2.66, so -32t = -113/2.66 which is the same value we found in number 2.

So solve for t

-32t = -42.48

Thus, t = 1.33 is the time at which the instantaneous and average velocity are equal.