Final Velocity of a car with a opposing decreasing drag force

[vbottas]

Homework Statement

How do I find the final velocity of a car moving initially at 30.63 m/s after it has moved 45m. Over the 45 meters, It has only 1 opposing force, drag force. Ignore the force of static friction propelling it foward. However, this drag force is decreasing since this car is slowing down so this drag force is not constant throughtout the 45m. Here is the data, mass of car is 1565m, density of air is 1.3, drag coefficient of the car is 0.20, cross sectional area is 8.71 m^2.

Relevant Equations

work equation?

Hello! How do I find the final velocity of a car moving initially at 30.63 m/s after it has moved 45m. Over the 45 meters, It has only 1 opposing force, drag force. Ignore the force of static friction propelling it foward. However, this drag force is decreasing since this car is slowing down so this drag force is not constant throughtout the 45m. Here is the data, mass of car is 1565m, density of air is 1.3, drag coefficient of the car is 0.20, cross sectional area is 8.71 m^2.

 

[erobz]

What is the drag force proportional to $v$ or $v^2$?

 

[vbottas]

v^2

 

[vbottas]

I feel like I need to use integrals somehow and relate it to work and kinetic energy

 

[erobz]

v^2

Ok, can you write Newtons 2nd law for the car during its motion?

 

[vbottas]

Ok, can you write Newtons 2nd law for the car during its motion?

yes, but how would we write it if the drag force isn't constant throughout the 45m displacement since velocity (a component of drag force) is decreasing through the interval?

 

[erobz]

yes, but how would we write it if the drag force isn't constant throughout the 45m displacement since velocity (a component of drag force) is decreasing through the interval?

Like a differential equation. $a = \frac{dv}{dt}$

 

[Steve4Physics]

Like a differential equation. $a = \frac{dv}{dx}$

Typo' alert! $a = \frac{dv}{dt}$

 

[erobz]

Typo' alert! $a = \frac{dv}{dt}$

Yeas, I was getting ahead of myself!

 

[vbottas]

so, F=ma, ∫Fddt=m∫dv/dt dt?

 

[erobz]

so, F=ma, ∫Fddt=m∫dv/dt dt?

No, not quite.

Just try writing the differential form of $\sum F=ma$ for this problem.

The drag force must be represented on the LHS.

 

[kuruman]

Please provide units with all the numbers you quote. What kind of car has mass 1565m?

 

[vbottas]

No, not quite.

Just try writing the differential form of $\sum F=ma$ for this problem.

The drag force must be represented on the LHS.

Ok this makes sense. So, ∫Fd dt= ma?

 

[collinsmark]

so, F=ma, ∫Fddt=m∫dv/dt dt?

A lot of us would like to help you out (as should be obvious by now, with all the replies), but you haven't even set up the basic equation yet.

Newton's second law is important. (Yes, you have mentioned that one.) But don't integrate that just yet. There's going to be a trick/shortcut you'll want to do first (more on that in due time).

There's also the drag force equation. And you haven't mentioned that one yet. So write that one down. The, for starters, incorporate the drag force equation into Newton's second law.

(Later, there's a little trick involving a = \frac{dv}{dt} = \frac{dv}{dt} \frac{dx}{dx} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}, but I'm getting ahead of myself too, and we'll get to that later.)

So to start, what's the drag force equation? What's Newton's second law?

 

[erobz]

Ok this makes sense. So, ∫Fd dt= ma?

No integrals at this point in the development of the ODE.

The LHS the the drag force. It’s not an integral as a “force”. It’s just “something” ( involving all this constants) times $v^2$

 

[vbottas]

Please provide units with all the numbers you quote. What kind of car has mass 1565m?

 

[vbottas]

A lot of us would like to help you out (as should be obvious by now, with all the replies), but you haven't even set up the basic equation yet.

Newton's second law is important. (Yes, you have mentioned that one.) But don't integrate that just yet. There's going to be a trick/shortcut you'll want to do first (more on that in due time).

There's also the drag force equation. And you haven't mentioned that one yet. So write that one down. The, for starters, incorporate the drag force equation into Newton's second law.

(Later, there's a little trick involving a = \frac{dv}{dt} = \frac{dv}{dt} \frac{dx}{dx} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}, but I'm getting ahead of myself too, and we'll get to that later.)

So to start, what's the drag force equation? What's Newton's second law?

Thank you. I appreciate each and everyone of y'all helping me out.
Drag force equation is Cd*A*p*v^2. therefore, CdApv^2=ma

 

[collinsmark]

Thank you. I appreciate each and everyone of y'all helping my out.
Drag force equation is Cd*A*p*v^2. therefore, CdApv^2=ma

Close, but shouldn't there be a \frac{1}{2} in there? And also, since the force is opposite the velocity, there needs to be a negative sign in there too.

 

[erobz]

Thank you. I appreciate each and everyone of y'all helping my out.
Drag force equation is Cd*A*p*v^2. therefore, CdApv^2=ma

Check the sign… the drag force is opposite the direction of motion. And express the RHS as $m\frac{dv}{dt}$( to start)

 

[vbottas]

Just got off work. Apologies for the late answer. Got it! Made the changes and corrected the RHS to m dv/dt

 

[vbottas]

And yes! there is a 1/2. good catch thank you. -(1/2)CdApv^2=m(dv/dt)

 

[erobz]

And yes! there is a 1/2. good catch thank you. -(1/2)CdApv^2=m(dv/dt)

Ok, now refer to the post #14 by @collinsmark to change independent variable from time $t$ to position $x$.

 

[collinsmark]

And yes! there is a 1/2. good catch thank you. -(1/2)CdApv^2=m(dv/dt)

OK. So far it looks good. Now the trick. Keep this little trick in your back pocket, not just for this problem, but for future problems. It is very useful when you want to convert to integrating over distance instead of time and vise versa. And the basic idea, with different variables that are related to each other, can help in other differential equations too, even if they don't involve time and distance specifically but are still related to each other.

So now we have

- \frac{1}{2} C_d A \rho v^2 = m \frac{dv}{dt}

The problem is that we don't really care about the time it takes the object to move. We are given that problem moves a specific distance. The distance is what's given, not the time. It would be much easier if we could integrate with respect to distance instead of time. Wouldn't it be nice if we could integrate with respect to dx instead of dt? For that, notice the following:

a = \frac{dv}{dt} = \frac{dx}{dx} \frac{dv}{dt} = \frac{dx}{dt} \frac{dv}{dx} = v \frac{dv}{dx}

Now see if you can use that in a creative and useful way, getting rid of any dt, and instead just having non-constant variables of dx, v, and dv.

 

[vbottas]

Ok, now refer to the post #14 by @collinsmark to change independent variable from time $t$ to position $x$.

Looking at the RHS of my equation, this could change it to m(v(dv/dt)) correct?

 

[collinsmark]

Looking at the RHS of my equation, this could change it to m(v(dv/dt)) correct?

That might be a typo, but not quite. Again, you're tying to get rid of the dts.

 

[vbottas]

I apologize. Yes that is a typo. m(vdv/dx) is what I meant to type

 

[collinsmark]

I apologize. Yes that is a typo. m(vdv/dx) is what I meant to type

Yes, that's right!

Now, incorporate that into your equation. As a special bonus, you should now be able to simplify it a bit before the next step.

 

[vbottas]

 

[vbottas]

Yes, that's right!

Now, incorporate that into your equation. As a special bonus, you should now be able to simplify it a bit before the next step.

taking one v out from both sides correct?

 

[collinsmark]

View attachment 340079

Yes, very nice. But do you see a way to simplify that a little? (Do you see any variables that cancel?)