Impossible Physics Questions???

[Death]

Okay wow. Our teacher just gave us a 3 random questions to answer and I can't seem to answer any of them.

1. Derive the formula for the Uniform Acceleration Formula :

Delta D = V2*Delta T - 1/2a*Delta T^2

V2^2 = V1^2 +2a*Delta D

2. An astronaut on the moon throws a wrench straight up at initial velocity of 4.0 m/s. Three seconds later, it falls back down at 0.8 m/s.

How long would it take the wrench to return to the position from which it was thrown??

Now, using the acceleration due to gravity on the moon, plus the two velocities and time (of 3 seconds) I was able to calculate the distance between the initial position and the apex: which is 4.8m.

For some reason, I can't solve the position question. I isolated T from:

Delta D = V1*Delta T + 1/2a*Delta T^2

This gives me the equation in quadratic formula but I keep getting answers like 3 seconds and 6.6 seconds.

Im supposed to get 5.0 seconds because our teacher gave us the answer. We just have to show the steps but no matter what, I can't get 5.0 seconds for Delta T.


HELP?!?!?!??!?

 

[GoldShadow]

Not sure how you're getting answers like 3 or 6.6.

Anyway, you don't even need to worry about things like "three seconds later" or "falls back down at 0.8m/s" or where the apex is (it's at exactly 5.0m, by the way, not 4.8m... not that it matters, you don't need it for this problem).

You need to find out when the wrench returns to the position it started from: zero. It started at d=0, and it will fall back down to d=0.

In other words, just set the original distance equation to zero, like so:

0 = (4.0m/s)t + (1/2)(-1.6 m/s^2)t^2

Use the quadratic formula to solve for t, and you get t=0 and t=5. Obviously, t=0 is when the wrench was first thrown up, so you want the second answer, t=5. That's when the wrench comes back down to where it was thrown from.

 

[Death]

Oh thanks a lot. I never even thought of the displacement being zero because I kept on using 4.8 or whatever :embarrass

Could you possibly help me with the derivation? I don't understand them at all.

Oh, one more question.

A girl throws a pebble into a well at 4.0 m/s. 2 seconds later it hits the water.
I figured out the depth to be 28 m/s and the average velocity to be 14 m/s.
Now I am asked to find how long it will take the pebble to go halfway.

I asked my teacher if i could just say 1 s because it took one second to reach the water so halfway there it was 1 s. But he said since it is acceleration I cannot do this????

 

[Lixivium]

Oh thanks a lot. I never even thought of the displacement being zero because I kept on using 4.8 or whatever :embarrass

Could you possibly help me with the derivation? I don't understand them at all.

You need calculus to derive it. Have you learned that yet?

Oh, one more question.

A girl throws a pebble into a well at 4.0 m/s. 2 seconds later it hits the water.
I figured out the depth to be 28 m/s and the average velocity to be 14 m/s.
Now I am asked to find how long it will take the pebble to go halfway.

First of all, m/s is a unit of velocity, not distance, second, I hope you know that the acceleration of gravity is really closer to 9.86 m/s, but sometimes people use 10 m/s as an approximation. (Which is what you did here).

I asked my teacher if i could just say 1 s because it took one second to reach the water so halfway there it was 1 s. But he said since it is acceleration I cannot do this????

He's right, think about it:

The pebble is accelerating as it falls. It will move faster during the second second than during the first. It will also travel a greater distance during the second second that during the first. So at one second, it has not yet traveled halfway down the well.

Use your acceleration forumla. You know the displacement, use it to solve for the time.

Define down the well to be the positive direction.

x = x0 + v0 * t + 1/2 a*t^2

v0 = 4 m/s (initial velocity)
x = 14 m (halfway down)
x0 = 0 m (initial position is at the top of the well)
a = 10 m/s^2 (approximate)

14 = 0 + 4t + 1/2 * 10 * (t^2)
5 * t^2 + 4 * t - 14 = 0
a = 5, b = 4, c = -14
Quadratic formula:
t = 1.22, -2.22 seconds

So the answer is the pebble fell halfway down at 1.22 seconds. (Approximate)

I hope that was right, it's been a while since I took physics class.

 

[Death]

Ok the original question was:

How soon after it is thrown does the pebble actually acquire the veolcity calculated in b) which equals 14 m/s. And the answer is 1s. I think I should be able to get 1s exactly?????

 

[Lixivium]

Ok the original question was:

How soon after it is thrown does the pebble actually acquire the veolcity calculated in b) which equals 14 m/s. And the answer is 1s. I think I should be able to get 1s exactly?????

Well yeah, if you're accelerating at 10 m/s, and you started with 4 m/s, then it takes exactly 1 second to get to 14 m/s (because you are now moving 10 m/s faster than when you started).

But the pebble is NOT halfway down the well when this happens!!!!

As I just showed you, it takes 1.22 seconds for the pebble to go halfway down. So when it is at 14 m/s, it has not yet fallen halfway.

 

[BlackFoxPariah]

We were doing stuff like this but I can't really help since my teacher said she won't give us problems using the quadratic formula...X_X

 

[Death]

So theres no need to use quadratic formula?? I just write :

V2 = V1 +at
14 m/s - 4 m/s / 9.8 m/s^2 = t
10 m/s / 9.8 m/s^2 = t
1.02 = t

YAY!!!

haha, thank you very much. Could I rely on you for future physics help????

 

[chaos_Leader]

Don't get me started on physics...



too late...

thats, not impossible, Thats basic kinematics.
If you want odd physics, try relativity:

time dilation, or the stretching or compressing of an object's relative time due to its velocity

here is the formula:

t = the change in time for the object in motion
t' = the change in time for the object at rest
v = the velocity of the object in motion
c = the speed of light ( roughly 3*10^8 m/s or 300,000,000 meters per second)

[the greek letter delta, the triangle, represents a change in a value, such as a change in time passed, not an actual number]

this is the formula from which the famous Twin Paradox is derived:

Lets say a man travels for 15 years relative to himself and a craft that moves at 87% of the speed of light. He needs to know how much time has passed relative to those who do not travel near the speed of light. (normally seconds is the standard unit of measure in physics but since both the original measure and answer are in years, and no one wants to deal with that many seconds, an exemption can be made)
it can be written like so:

v = 0.87c
t = 15y
c = 8*10^6 m/s
t' = ?

t' = 15y / Sqrt[ 1 - (0.87c^2) / c^2 ]

You want to deal with that square root first. now you could drag it out and end with a very big number for your velocity, but you can avoid that headache easy:

(0.87c)^2 = (0.87^2) (c^2)

now under the square root symbol you have:

1 - (0.87^2 * c^2) / (c^2)

more headaches are avoided when c^2 cancels itself out

1 - (0.87^2 * c^2) / (c^2)

leaving you with:

1 - .87^2

that can be calculated to:

~ 0.24

now the square root of 0.24

Sqrt[0.24] = 0.48

This leaves a far simpler formula now:

t' = 15y / 0.48

t' = 31.25y

so about 31 and a quarter years have passed in the outside world while the man traveling at 87% of the speed of light only experienced 15 years.

This formula has been tested with massive particle accelerators and two atomic clocks, one on a plane and one on the ground, this formula of dime dilation holds true.


That ends this post's lesson on special relativity. I hope your brain isn't too scrambled.