Question

Please explain your responses as much as you possibly can. It would really be appreciated :)

1. On the moon, the acceleration due to gravity is 1.6 m/sec^2.

a. If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec later?

b. How far below the point of release is the bottom of the crevasse?

c. If instead of being released from rest, the rock is thrown into the crevasse from the same point with a downward velocity of 4m/sec, when will it hit the bottom and how fast will it be going when it does?

2.a. With what velocity will you hit the water if you step off from a 10-m diving platform?
b. With what velocity will you hit the water if you dive off the platform with an upward velocity of 2m/sec?

3. The function

f(x) = {x, 0 is less than or equal to x <1
{0, x = 1

is zero at x=0 and at x=1. Its derivative is equal to 1 at every point between 0 and 1, so f' is never zero between 0 and 1, and the graph of f has no tangent parallel to the chord from (0,0) to (1,0). Explain why this does not contradict the Mean Value Theorem.

THANK YOU SO MUCH!!!

Answer 1

1)

a)We start by knowing that the acceleration of the falling object is 1.6 m/s^2 on the moon and it hits the bottom of the crevasse 30 seconds after being released from rest.

Recall that we can think of the second derivative of a function as the acceleration of the function. Since acceleration due to gravity is essentially a constant we just have:

y''(t)=-1.6 (y'' means second derivative here incase the notation is foreign and acceleration is downward so i made it negative)

Integrating once with respect to time yields the velocity formula:

y'(t)=-1.6t+c

where c is the integration constant. physically it is an initial velocity. But since we released the object from rest c=0 since it had no initial velocity at t=0.

Now:

y'(t)=-1.6t

To find the velocity when it hits the ground at t=30 just plug in 30 and find:

y'(t)=-48

So the object was falling at -48 m/s on when it hits the bottom.

b) how far is the bottom?

Integrating again yields us the position equation:

y(t)=-0.8t^2 +h where h is the integration constant physically interpreted to be the initial height.

since at t=30 it hits the bottom we can plug in t=30 and solve for h ( because y(30)=0)

0=-0.8(30)^2 + h

solving for h:

h=720 meters.

c) What if the initial velocity is 4 m/s instead of zero. How long will it take to hit the bottom and how fast will it be going?

So going back to the general velocity formula and plugging in c=- 4. (neg because directed downward)

y'(t)=-1.8t-4

In order to find how long it took to reach the bottom we need to integrate to find the position (note the initial position is still 720):

y(t)=-0.8 t^2 -4t +720

We need to find when at what t does y(t)=0?

0=-0.8 t^2-4t +720

Using the quadratic formula we find (negative time doesn't make sense in the context of the problem so the positive solution is the only applicable one):

t= 27.06 (i rounded here so its not exact)

plugging this into the velocity equation:

y'(27.06)=-1.6(27.06)-4 = -47.3 m/s (i rounded again)

2)

a)What velocity will you hit the water if you fall off a 10m diving board?

Acceleration due to gravity on earth is -9.8 m/s^2. The acceleration equation is:

y''(t)=-9.8

Integrating (velocity formula):

y'(t)=-9.8t + c (c= 0 here)

So:

y'(t)=-9.8t

Integrating again(position formula):

y(t)=-4.9t^2 + h (h=10 as given in the problem)

plugging in h=10

y(t)=-4.9t^2 + 10

At what time does the position equal 0?

0=-4.9t^2 + 10

Solving for t (positive only again here... negative time doesn't mean anything in context of problem):

t=1.43 (rounding)

Plugging into velocity formula:

y'(1.43)=-9.8(1.43)=-14 m/s (i rounded here again)

b) with what velocity will you hit the water if you dive off with and upward velocity of 2 m/s?

So the velocity formula now reads(c=2):

y'(t)=-9.8t+2

Integrating to get the position formula (initial height is still 10):

y(t)=-4.9t^2+2t+10

Using the quadratic formula to solve when y(t)=0:

0=-4.9t^2+2t+10

t=1.65 (negative answer doesn't make sense in context of problem; rounding)

Using this value for the velocity equation:

y'(1.65)=-9.8(1.65)+2= -14.17 m/s

3) suppose:

is zero at x=0 and x=1. Its derivative is equal to 1 for every point between 0 and 1. Explain why this does not violate the mean value theorem.

By definition the mean value theorem applies to smooth continuous functions. But our function has a discontinuity at x=1. Therefore the mean value theorem does not apply. Therefore it does not violate the mean value theorem.

Hope this was helpful