Finding Distance: Penny in a Wishing Well

[mogibb1]

Homework Statement

Aunt Minnie drops a penny into a wishing well and it falls for 3 seconds before hitting the water. How far down is the water surface?

Homework Equations

d = 5t²

d = 5(3)²

d = 45 m

Followup question: Aunt Minnie didn't get her wish so she goes to a deeper well and throws in a penny straight down into it at 10 m/s. How far does this penny go in 3 seconds?

Isn't this the same question? It's still travels only 3 seconds. I'm not sure how to figure this one out.

The Attempt at a Solution

 

[skeptic2]

In the followup question the penny starts with an initial velocity of 10 m/s but the acceleration is the same. Perhaps you could work it as two separate problems. How far will the the penny travel at 10 m/s plus how far will the penny fall at the rate of 5t^2? Can you combine the two solutions into one formula?

 

[mogibb1]

At 10 m/s the distance will be:

d=5(1)²

d=5m

for 20 m/s the distance will be:

d=5(2)²

d=20m

for 30 m/s the distance will be:

d=5(3)₂

d=45m

Because she throws it an initial velocity of 10 m/s, will it still be 10 m/s for the first second, 20 m/s for second second, and 30 m/s for third second?

If that the case then the answer would be 75m because the between the 0s and 1s would be 5m and between 1s and 2s the distance would be 5m and between 2s and 3s it would be 20m and at the 3s it would be 45m. Add them up and get 75m. Does this make any sense?

 

[Dickfore]

What is the formula for distance traveled during uniformly accelerated motion with some initial velocity?

 

[mogibb1]

d=1/2gt² or

d=5t²

 

[Dickfore]

No, this is if the initial velocity is zero.

 

[mogibb1]

Oh. I don't know what the formula is then.

 

[Dickfore]

Ok, do you know the formula for the instantaneous velocity during a uniformly accelerated motion?

 

[mogibb1]

V=V₀+at

This may not be it, but I'm trying to make sense of this stuff. My book doesn't give very much help.

 

[Dickfore]

Yes. So, let's derive the formula for the distance traveled when there is initial velocity in the following shortcut way:

Imagine that the object was thrown a little time before \tau so that when it passes at the top of the well, it has exactly velocity v_{0}. From the formula you had posted (with v_{0} = 0 in it and v = v_{0}, think about it!), we would have:

<br /> v_{0} = a \tau<br />

by this time, the object had displaced by:

<br /> s_{0} = \frac{1}{2} a t^{2}<br />

Next, let us turn to the part of the motion from the time when it passes by the top of the well. In this case, the object has some initial velocity. Let us see how much it displaces after a time t had passed. The total time it had traveled is t + \tau. During this time, it displaced by:

<br /> \tilde{s} = \frac{1}{2} a (t + \tau)^{2}<br />

During the period that we are interested in, however, it displaced by only (make a sketch to verify!):

<br /> s = \tilde{s} - s_{0}<br />

Using the above formulas and the binomial formula, we get:

<br /> s = \frac{1}{2} a (t + \tau)^{2} - \frac{1}{2} a \tau^{2}<br />
<br /> s = \frac{1}{2} a \left[(t + \tau)^{2} - \tau^{2}\right]<br />
<br /> s = \frac{1}{2} a \left(t^{2} + 2 t \tau + \tau^{2} - \tau^{2} \right)<br />
<br /> s = a \tau t + \frac{1}{2} a t^{2}<br />

Finally, we need to eliminate the "fine-tuning" parameter \tau and expresses it through the information that we really have, namely the initial velocity. For this, use the first equation. Then, the formula for displacement becomes:

<br /> s = v_{0} t + \frac{1}{2} a t^{2}<br />

 

[mogibb1]

Ok so it would be:


Initial Velocity Gravity
| Time | Time
| | | |
X=10 m/s (3)+1/2(10)3²

x=30 m/s +5(9)

x=75m