Variable Acceleration - Clueless

In summary, The problem is labeled Variable Acceleration.The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by the expression Ax = -3.00 Vx^2. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to one-half of it's initial value.The professor has given us the answers so the answer is .222 seconds. I am not able to attempt a solution I can only restate the variables in more readable terms.Final speed (Vf) = .75 Initial speed (
  • #1
ssjcory
11
0

Homework Statement


The problem is labeled Variable Acceleration.

The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by the expression Ax = -3.00 Vx^2. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to one-half of it's initial value.

Homework Equations


I'm not sure (I suspect that the equations of motion are used?)

The Attempt at a Solution


The professor has given us the answers so the answer is .222 seconds.
I am not able to attempt a solution I can only restate the variables in more readable terms.

Final speed (Vf) = .75
Initial speed (V0) = 1.50
Initial time (T0) = 0
Initial position(X0) = 0
Equation for var accel (Ax) = -3.00 Vx^2
Final time (Tf) = ? this is the goal

I know the problem has to do with integration but I'm not sure where to start.
Can someone help me out?

Thanks,
Cory
 
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  • #2
ssjcory said:

Homework Statement


The problem is labeled Variable Acceleration.

The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by the expression Ax = -3.00 Vx^2. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to one-half of it's initial value.


Homework Equations


I'm not sure (I suspect that the equations of motion are used?)


The Attempt at a Solution


The professor has given us the answers so the answer is .222 seconds.
I am not able to attempt a solution I can only restate the variables in more readable terms.

Final speed (Vf) = .75
Initial speed (V0) = 1.50
Initial time (T0) = 0
Initial position(X0) = 0
Equation for var accel (Ax) = -3.00 Vx^2
Final time (Tf) = ? this is the goal

I know the problem has to do with integration but I'm not sure where to start.
Can someone help me out?

Thanks,
Cory

Have you learned differential equations? You'll definitely need to set one up to solve this.
 
  • #3
Curious3141 said:
Have you learned differential equations? You'll definitely need to set one up to solve this.

I did differentiation and integration in calc 1 but I'm really not sure how to apply it here.
 
  • #4
ssjcory said:

Homework Statement


The problem is labeled Variable Acceleration.

The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by the expression Ax = -3.00 Vx^2. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to one-half of it's initial value.
...

Can someone help me out?

Thanks,
Cory
Hopefully you know that [itex]\displaystyle a=\frac{dv}{dt}\,.[/itex]

Substituting (-3.00)v2 in for a gives:
[itex]\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.[/itex]​

Divide both sides by v2 and integrate with respect to time.
 
  • #5
SammyS said:
Hopefully you know that [itex]\displaystyle a=\frac{dv}{dt}\,.[/itex]

Substituting (-3.00)v2 in for a gives:
[itex]\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.[/itex]​

Divide both sides by v2 and integrate with respect to time.

I am definitely missing something big here.

I see that [itex]\displaystyle a=\frac{dv}{dt}\,.[/itex]

and I know [itex]\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.[/itex]
because we are given [itex]\displaystyle a=(-3.00)v^2[/itex]

As you said divide each side by v^2 (I guess so that we have all the variables on the right.
So we get [itex]\displaystyle -3.00= \frac{dv}{(dt)(v^2)}\,[/itex]

It can be rewritten to have dt * v^-2 on top and dt on the bottom.

So now -3 = (dv*v^-2) / dt ... I am given initial and final velocity which I know have to go into this equation somehow. Initial = 1.5, final=.75

I am completely lost again :(
By virtue of knowing the answer I know that 1.5^-2 + .75^-2 ... yields the answer ... but I have no true idea why because I don't understand where that dv/dt went and I didn't really integrate (aka add one to the exponent and divide by that exponent)

I know I am not seeing something clearly. It has been a few years since calculus but I feel like its not the physics or the calculus that are messing me up, but where they intersect.

Thanks for your time,
Cory
 
  • #6
ssjcory said:
I am definitely missing something big here.

I see that [itex]\displaystyle a=\frac{dv}{dt}\,.[/itex]

and I know [itex]\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.[/itex]
because we are given [itex]\displaystyle a=(-3.00)v^2[/itex]

As you said divide each side by v^2 (I guess so that we have all the variables on the right.
So we get [itex]\displaystyle -3.00= \frac{dv}{(dt)(v^2)}\,[/itex]

It can be rewritten to have dt * v^-2 on top and dt on the bottom.
...

Thanks for your time,
Cory
OK! You have [itex]\displaystyle -3.00= v^{-2}\frac{dv}{dt}\,.[/itex]

Now integrate:
[itex]\displaystyle \int-3.00\,dt= \int v^{-2}\frac{dv}{dt}dt[/itex]

[itex]\displaystyle =\int v^{-2} dv [/itex]​

Proceed on.
 
  • #7
SammyS said:
OK! You have [itex]\displaystyle -3.00= v^{-2}\frac{dv}{dt}\,.[/itex]

Now integrate:
[itex]\displaystyle \int-3.00\,dt= \int v^{-2}\frac{dv}{dt}dt[/itex]

[itex]\displaystyle =\int v^{-2} dv [/itex]​

Proceed on.
Wouldn't the integration of -3 be -3X(or -3T in this circumstance?)

And on the right side v^-2 should become v^(n+1)/n+1 ... Yielding (v^-1)/1 ... or v^-1 ?

So -3T = v^-1? And plug in the final and then initial velocity?
-3T = 1.5^-1 === -.222
-3T = .75^-1 === -.444

This is very confusing to me would it be -.444(equation of time with final velocity) minus -.222(equation of time with initial velocity) ... = -.222 ... Do I just disregard the negative because it is the time? Or have I completely screwed this up?

Thanks!
Cory
 
  • #8
ssjcory said:
Wouldn't the integration of -3 be -3X(or -3T in this circumstance?)

And on the right side v^-2 should become v^(n+1)/n+1 ... Yielding (v^-1)/1 ... or v^-1 ?

So -3T = v^-1? And plug in the final and then initial velocity?
-3T = 1.5^-1 === -.222
-3T = .75^-1 === -.444

This is very confusing to me would it be -.444(equation of time with final velocity) minus -.222(equation of time with initial velocity) ... = -.222 ... Do I just disregard the negative because it is the time? Or have I completely screwed this up?

Thanks!
Cory

The indefinite integral of [itex]v^{-2}[/itex] wrt v is [itex]-v^{-1} + C= -\frac{1}{v} + C[/itex]. You missed out the negative sign in yours (remember that the "n+1" here is MINUS one) and the constant of integration.

I've always found it better to use definite integration on both sides and incorporate the initial conditions that way. This way you don't have to bother to solve for the constant of integration.

So rewrite as:

[tex]\int_0^t (-3)dt = \int_{1.5}^v v^{-2}dv[/tex]

and here, you're basically saying that at t=0, v = 1.5 m/s (lower bound). At an unknown time t, the velocity is just v (upper bound). Although this is not the best way to write the integral from a purist mathematical perspective, it's very common in physics, and hence is acceptable.

Now integrate and impose the bounds:

[tex]-3t = -\frac{1}{v} + \frac{1}{1.5}[/tex]

See how the bounds were incorporated there. On the left hand side, the term for t=0 vanishes. On the right hand side, you need to subtract off the -1/v expression at v = 1.5, but subtracting a negative is like adding it (minus*minus = plus).

All you're required to do now is to find t when v is half the initial velocity, or 0.75 m/s.
 
  • #9
Curious3141 said:
The indefinite integral of [itex]v^{-2}[/itex] wrt v is [itex]-v^{-1} + C= -\frac{1}{v} + C[/itex]. You missed out the negative sign in yours (remember that the "n+1" here is MINUS one) and the constant of integration.

I've always found it better to use definite integration on both sides and incorporate the initial conditions that way. This way you don't have to bother to solve for the constant of integration.

So rewrite as:

[tex]\int_0^t (-3)dt = \int_{1.5}^v v^{-2}dv[/tex]

and here, you're basically saying that at t=0, v = 1.5 m/s (lower bound). At an unknown time t, the velocity is just v (upper bound). Although this is not the best way to write the integral from a purist mathematical perspective, it's very common in physics, and hence is acceptable.

Now integrate and impose the bounds:

[tex]-3t = -\frac{1}{v} + \frac{1}{1.5}[/tex]

See how the bounds were incorporated there. On the left hand side, the term for t=0 vanishes. On the right hand side, you need to subtract off the -1/v expression at v = 1.5, but subtracting a negative is like adding it (minus*minus = plus).

All you're required to do now is to find t when v is half the initial velocity, or 0.75 m/s.

So I was just a tad bit off for the integration (meant to put the negative in front of the v^-1)

So once we have the basic form
-3T = -v^-1

Which I think I was able to get to.
You have to add the initial velocity since that is where the equation starts?
I'm guessing that's something you just have to think about and know to do.

After that I plugged in .75 into v and ended up with t = .222 which is the right answer :)
I double checked by plugging in the initial values (T=0) and (V=1.5) to make sure that both sides of the equation balanced. They did so I knew it was correct :D


I have one last question for ya.
The dv or dt don't really matter do they? Are they just an indicator of what we are integrating in respect to?

Thanks so much you are a life saver. I know one of these questions is going to be on the exam (Only 8 questions on it!)

Thanks,
Cory
 
  • #10
ssjcory said:
So I was just a tad bit off for the integration (meant to put the negative in front of the v^-1)

So once we have the basic form
-3T = -v^-1

Which I think I was able to get to.
You have to add the initial velocity since that is where the equation starts?
I'm guessing that's something you just have to think about and know to do.

After that I plugged in .75 into v and ended up with t = .222 which is the right answer :)
I double checked by plugging in the initial values (T=0) and (V=1.5) to make sure that both sides of the equation balanced. They did so I knew it was correct :D

It's unclear what you're doing, so I can't tell if you're really doing the right thing. I don't understand what you mean by "basic form" here. The velocity at time t is *not* given by "-3T = -v^-1" because you haven't taken into account the fact that the initial velocity is 1.5 m/s. You either have to consider the constant of integration (and then solve for it using the initial conditions), or you have to set up the definite integral as I did. Either way, what I posted is the actual relation between v and t.

I have one last question for ya.
The dv or dt don't really matter do they? Are they just an indicator of what we are integrating in respect to?

Thanks so much you are a life saver. I know one of these questions is going to be on the exam (Only 8 questions on it!)

Thanks,
Cory

Of course they matter! They're the variables of integration. Manipulating them algebraically to bring them on either side of the equation (called separating the variables) is just a way to solve this very simple differential equation.
 

1. What is variable acceleration?

Variable acceleration is the change in velocity over time. This means that an object's speed and/or direction is constantly changing, rather than remaining constant.

2. How is variable acceleration different from constant acceleration?

Constant acceleration is when an object's velocity changes by the same amount in the same direction over equal intervals of time. Variable acceleration, on the other hand, means that the object's velocity is changing at different rates or in different directions at different times.

3. What causes variable acceleration?

Variable acceleration can be caused by forces acting on an object, such as gravity, friction, or air resistance. These forces can change the object's velocity, resulting in variable acceleration.

4. How is variable acceleration calculated?

Variable acceleration can be calculated by finding the change in velocity over a specific time interval. This is done by dividing the change in velocity by the change in time, using the formula a = (vf - vi)/t.

5. What are some real-life examples of variable acceleration?

Some real-life examples of variable acceleration include a car accelerating and decelerating in traffic, a roller coaster speeding up and slowing down as it goes through loops and turns, and a skydiver falling towards the ground while adjusting their body position to change their speed and direction.

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