Velocity equation and distance problems

[Krokodrile]

Homework Statement

1.- A steel ball is released from rest in a container of oil. Its downward acceleration is a = 2.4 - 0.6v inch/s^2. what is the ball's downward velocity 2 seconds after it has been released?
2.-What is the distance that the ball falls in the first 2 seconds after being released?

Relevant Equations

dv/dt

I try to resolved with my knowlegde of the dynamic class:

Acceleration is known to be the derivative of velocity with respect to time.
a = dv / dt; so that dv = a dt
Replace: dv = (2.4 - 0.6v inch/s^2) dt
Then: dt = dv / (2.4 - 0.6v inch/s^2); we integrate t between 0 and 2; v between 0 and v
t = int [dv / (2.4 - 0.6v inch/s^2)] = - Ln (2.4 - 0.6v inch/s^2) / 0.6; we must clear v
Ln (2.4 - 0.6v inch/s^2) = - 0.6 t
2.4 - 0.6v inch/s^2 = e^(-0.6 t)
Therefore v = [2.4 + e ^ (- 0.6 t)] / 0.6
Since the initial velocity is zero, we must add a constant to the equation for the final velocity.
t = 0; implies v = 0 = [2.4 + e ^ (- 0.6 t ) / 0.6 + k = (2.4 + 1) / 0.6 + k = 0
So k = - (2.4 + 1) / 0.6
Finally: v = [ 2.4 + e ^ (- 0.6 t)] / 0.6 - (2.4 + 1) / 0.6

 

[haruspex]

Finally: v = [ 2.4 + e ^ (- 0.6 t)] / 0.6 - (2.4 + 1) / 0.6

Looks like you dropped a factor somewhere. In your final equation the two 2.4/0.6 terms cancel.
Check your equation by differentiating.

 

[PeroK]

Homework Statement:: 1.- A steel ball is released from rest in a container of oil. Its downward acceleration is a = 2.4 - 0.6v inch/s^2. what is the ball's downward velocity 2 seconds after it has been released?
2.-What is the distance that the ball falls in the first 2 seconds after being released?
Relevant Equations:: dv/dt
Finally: v = [ 2.4 + e ^ (- 0.6 t)] / 0.6 - (2.4 + 1) / 0.6

So, you have ##a = \alpha - \beta v##?

And, it would be the same set of solution steps whatever ##\alpha## and ##\beta## are given?

You could check the answer by differentiating.

 

[vela]

It's generally easier to spot mistakes when you stick with symbols rather than using numerical constants. Plus it's usually less writing.

 

[kuruman]

Since the initial velocity is zero, we must add a constant to the equation for the final velocity.

If you do the definite integral ##\int_0^v\frac{dv}{a-bv}## correctly, the integration constant is automatically taken care of and you don't have to add anything. The other side of the equation, of course, should be ##\int_0^t dt.##

 

[Krokodrile]

The teacher show``´ s the answer but no the way for resolve, so i try to solved with the integration v to 0 dv/a-bv and i got the corret answer: 2.795 (well, close to the answer). Thank you very much for you help ;)

 

[PeroK]

The teacher show``´ s the answer but no the way for resolve, so i try to solved with the integration v to 0 dv/a-bv and i got the corret answer: 2.795 (well, close to the answer). Thank you very much for you help ;)

That answer looks wrong. You could simulate the motion using a simple computer program or spreadsheet.

 

[Krokodrile]

That answer looks wrong. You could simulate the motion using a simple computer program or spreadsheet.

The answer is in cm/s^2, i converted the english units in international units

 

[PeroK]

The answer is in cm/s^2, i converted the english units in international units

Ah, so that was the answer to question 1). You ought to pay attention to units. An answer of 2.795 is meaningless in physics. Is that 2.795 inches/second?

 

[Krokodrile]

Ah, so that was the answer to question 1). You ought to pay attention to units. An answer of 2.795 is meaningless in physics. Is that 2.795 inches/second?

2.795 cm/s^2 after 2 seconds, and the second answer with integration and derivates help of the first question is 3.34 cm after 2 seconds

 

[PeroK]

2.795 cm/s^2 after 2 seconds, and the second answer with integration and derivates help of the first question is 3.34 cm after 2 seconds

Those numbers are correct using the units specified in the original question. You can't just swap from inches to cm.

 

[Krokodrile]

Those numbers are correct using the units specified in the original question. You can't just swap from inches to cm.

yes, that's my mistake. But I am very happy for got the correct method. PD: any recomendation for software simulation for this kind of problems?

 

[PeroK]

yes, that's my mistake. But I am very happy for got the correct method. PD: any recomendation for software simulation for this kind of problems?

If we have an acceleration ##a(t) = a_0 - bv(t)##, then the basic idea is:

We start with ##d = 0, v = 0, a = a_0## and then take a small time increment ##\Delta t## to get: $$v(\Delta t) \approx a_0 \Delta t, \ \ d(\Delta t) \approx \frac 1 2 a_0 (\Delta t)^2 \ \ \text{and} \ \ a(\Delta t) \approx a_0 - bv(\Delta t)$$ We can then use these new values for ##a## and ##v## to estimate the next increment:
$$d(t + \Delta t) \approx d(t) + v(t)\Delta t + \frac 1 2 a(t) (\Delta t)^2, \ \ v(t + \Delta t) \approx v(t) + a(t)\Delta t, \ \ a(t) \approx a_0 - bv(t)$$ And then use these interations to model the motion. E.g. we could set ##\Delta t = 0.1s## and apply ##20## interations to get from ##0s## to ##2s##.

You can do this easily on a spreadsheet or a simple Python program. I guess you don't get taught this sort of thing at school or university?

PS Integration takes this process of increments to the limit as ##\Delta t \rightarrow 0##.

 

[Krokodrile]

If we have an acceleration ##a(t) = a_0 - bv(t)##, then the basic idea is:

We start with ##d = 0, v = 0, a = a_0## and then take a small time increment ##\Delta t## to get: $$v(\Delta t) \approx a_0 \Delta t, \ \ d(\Delta t) \approx \frac 1 2 a_0 (\Delta t)^2 \ \ \text{and} \ \ a(\Delta t) \approx a_0 - bv(\Delta t)$$ We can then use these new values for ##a## and ##v## to estimate the next increment:
$$d(t + \Delta t) \approx d(t) + v(t)\Delta t + \frac 1 2 a(t) (\Delta t)^2, \ \ v(t + \Delta t) \approx = v(t) + a(t)\Delta t, \ \ a(t + \Delta t) \approx a_0 - bv(t)$$ And then use these interations to model the motion. E.g. we could set ##\Delta t = 0.1s## and apply ##20## interations to get from ##0s## to ##2s##.

You can do this easily on a spreadsheet or a simple Python program. I guess you don't get taught this sort of thing at school or university?

PS Integration takes this process of increments to the limit as ##\Delta t \rightarrow 0##.

Im impressed with that explanation and not...4th semester mechanic engineering only have basic knowlegdes in c`++.