Mean Value Theorem

Question

If a rock is dropped from a height of 5353 ft, its poistion tt seconds after it is dropped until it hits the ground is given by the function s(t)=−16t2+53s(t)=-16t2+53.

What time after the rock is thrown will its instantaneous velocity be equal to its average velcity? (Apply the Mean Value Theorem)

Answer 1

1. The height of the rock is given by: s(t) = -16t² + 53
2. This function is a simple quadratic, which is a polynomial, and all polynomials are continuous everywhere (-∞, ∞). Polynomials are also differentiable on all open intervals.
3. This means that we can choose any interval [a, b] that we want, where t starts at a and then ends up at b. Because it is a polynomial it will be continuous on [a, b] and differentiable on (a, b), which means we can invoke the mean value theorem.
4. The mean value theorem basically says that for a continuous and differentiable function defined over an interval [a, b], there exists c, where a < c < b, such that s'(c) = (s(b) - s(a))/(b - a)
5. In plain English, this means there is a point in time between a and b where the instantaneous velocity (tangent) equals the average velocity over the whole interval (secant).
6. In this example, a would be t = 0 when we drop the rock and t = b would be the time at which it hits the ground.
7. So the mean value theorem tells us that such a value t=c exists, but it doesn't tell us what it is or how to find it. We have to be a little more creative to find that. Our basic strategy should be to first find out what the average velocity is over the interval and then find out when the velocity equation, v(t) (the derivative of s(t)), equals that value.
8. Step 1 - Find how long it takes for the rock to hit the ground
9. 0 = -16t² + 53
10. logically, we are saying at what point t does the height = 0?
11. 16t^2 = 53
12. t^2 = 53/16
13. t = +-√(53/16)
14. a negative t value would be outside of the domain, so choose t = √(53/16), which is approximately 1.820 seconds
15. so a = 0, b = √(53/16), f(a) = 53, f(b) = 0
16. Step 2 - Calculate the average velocity over the interval
17. velocity = Δs/Δt = (f(b) - f(a))/(b - a) = (0 - 53)/(√53/16 - 0) = -53/1.82 = -29.12 ft/s
18. the negative sign indicates that it is falling downwards
19. Step 3 - See at which point in time the velocity function reaches this average velocity
20. s(t) = -16t² + 53
21. v(t) = s'(t) = -32t
22. velocity is the derivative of position
23. we are just using the power rule and sum rule here
24. -29.12 = -32t
25. set the velocity equation equal to the velocity we want so we can solve for t
26. this tells us when that exact velocity is reached
27. t = -29.12/-32 = 0.91 s
28. So the rock should take 1.82 seconds to hit the ground and it should be going as fast as its average speed over the whole interval approximately 0.91 seconds after you let it go