Simple velocity/acceleration/distance equation-but is mastering physics wrong?

• marewrath
In summary, the rock climber threw two stones vertically downward 1.0 apart and observed that they caused a single splash. The initial speed of the first stone was 1.6 . The speed of the first stone as it hit the water was vf=33.94 and the speed of the second stone as it hit the water was vf=53.619.
marewrath
simple velocity/acceleration/distance equation--but is mastering physics wrong?

A rock climber stands on top of a 57 -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 apart and observes that they cause a single splash. The initial speed of the first stone was 1.6 .

i will give you time, they both hit the water 3.3 seconds after the first stone is released (this both mastering physics and I agreed upon)

1. the initial speed of the second stone
2. the speed of the first stone as it hits the water
3. the speed of the second stone as it hits the water

so actually completed the questions, but the answers I got were different from the mastering physics answers--and I can't figure out how they got them. so if someone can scan these problems real quick and shoot their answers, id like to see if its mastering physics or me that is wrong.

Hi marewrath, welcome to PF.
Show your calculations, so that we can check it.

okay - i guess I jut wanted to check other peoples answers against mine versus masteringphysics, but here goes:

1. initial speed of the second stone:

57=vi+.5(9.8)(2.3)^2
57=vi+25.921
vi=31.079

2. speed of second stone as it hits the water

vf = 31.079 + 2.3(9.8)
vf=53.619
(of course, this is dependent on the answer from the last question)3. the speed of the first stone as it hits the water
vf=1.6+3.3(9.8)
vf=33.94

marewrath said:
A rock climber stands on top of a 57 [furlong? light year?] -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 [hours? orbital periods of Venus?] apart and observes that they cause a single splash. The initial speed of the first stone was 1.6 [c? knots? ] .
The point I'm trying to make by inserting some ridiculous, unrealistic units in the places where you left them out is that units are very important and you should always always include them in both the problem statement and in every step of your calculation. We don't tell you that just for the sake of being pedantic. Having the units in there helps you make certain your equations are dimensionally consistent (i.e. type of quantity on lefthand side = type of quantity on righthand side). If your units don't work out in this way, then you know you've made a mistake, so units are a valuable diagnostic. If you had done that, you would have caught the error you made below

marewrath said:
okay - i guess I jut wanted to check other peoples answers against mine versus masteringphysics, but here goes:

1. initial speed of the second stone:

57=vi+.5(9.8)(2.3)^2
Well, here is the problem. The equation is supposed to be:

d = vit + (1/2)at2

Without that factor of t in there, your equation is not dimensionally consistent, because you have:

[length] = [velocity]

which is nonsense. That's why you should always include the units in the calculation steps and make sure they work out.

i tried that originally and still wasnt getting the correct answers, but the ones without t times vi were closer so i figured that was probably more right and that t pertained to how long it was just going for that velocity... shrug

marewrath said:
i tried that originally and still wasnt getting the correct answers, but the ones without t times vi were closer so i figured that was probably more right

There is only one equation that is correct here, and it is the one I quoted for you. I would have hoped that you would have been convinced by the fact that your equation is not dimensionally consistent, rendering it meaningless! I would also hope that it would be clear to you that you can't just arbitrarily change the form of an equation and expect it to still be correct.
marewrath said:
and that t pertained to how long it was just going for that velocity... shrug

Nope. First of all, it's only "going at that [initial]" velocity for a single instant (i.e. for a time interval of zero width), so that doesn't make much sense. Second, If 't' in the first term of the equation was supposed to have meant something different from 't' in the second term of the equation, then we would have used a different symbol in order to make that explicitly clear. Same symbol means same quantity. In this case, t is meant to be the total time spent falling. To make this clearer:

vit - is the distance the rock *would* travel in time t if it started from speed vi and if it were NOT accelerating.

(1/2)at2 - is the distance the rock would travel in time t if it started from *rest* and were accelerating at 'a.'

In the most general case, you add these two distances together to get the total distance travelled.

hey hey! I wasn't saying I didnt believe you, or that I thought you were wrong at all, you are obviously correct; I was essentially saying that I confused myself early on, and thus began doing things wronger as I worked

marewrath said:
hey hey! I wasn't saying I didnt believe you, or that I thought you were wrong at all, you are obviously correct; I was essentially saying that I confused myself early on, and thus began doing things wronger as I worked

Alright I understand. Sorry if I seemed...animated. I just wanted to ensure you understood where you went wrong. Could you post what you got using the correct equations? It's possible that you just had a calculation error in there somewhere. I've done it and check against mine...

1. How do you calculate velocity using the simple equation?

The simple equation for velocity is v = d/t, where v is velocity, d is distance, and t is time. To calculate velocity, divide the distance traveled by the time it took to travel that distance.

2. What is the difference between velocity and acceleration?

Velocity is the rate at which an object changes its position. It includes both speed and direction. Acceleration, on the other hand, is the rate of change of an object's velocity. It measures how quickly the velocity is changing.

3. How do you use the simple equation to calculate acceleration?

The simple equation for acceleration is a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. To calculate acceleration, subtract the initial velocity from the final velocity and divide by the time interval.

4. Can mastering physics be wrong in its calculations?

While mastering physics is a helpful tool for practicing and understanding physics concepts, it is not infallible. There may be instances where the program may give incorrect calculations, so it is important to double-check your work and consult with a teacher or tutor if you are unsure.

5. How can I use the simple equation to calculate distance?

The simple equation for distance is d = vt, where d is distance, v is velocity, and t is time. To calculate distance, multiply the velocity by the time interval. This equation assumes a constant velocity, so it may not be accurate for more complex situations where the velocity is changing.

• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
22
Views
678
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
23
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
53
Views
4K
• Introductory Physics Homework Help
Replies
5
Views
7K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K