Final Velocity of a car with a opposing decreasing drag force

[collinsmark]

View attachment 340081

Perfect. Now integrate both sides of the equation. (This will involve two integrals, one on each side of the equation.)

 

[erobz]

should I plug in known values to the left side?

Always try to work in variables till the end. Calculate as last step.

 

[vbottas]

Alway try to work in variables till the end.

ok! I was just worried taking the integral this many constants labeled as letters may get a bit confusing. But I'll trust you on this!

 

[erobz]

ok! I was just worried taking the integral this many constants labeled as letters may get a bit confusing. But I'll trust you on this!

The constants come outside the integral.

 

[vbottas]

Perfect. Now integrate both sides of the equation. (This will involve two integrals, one on each side of the equation.)

the bounds on the left side is 0 to 45 correct? what would my right side bounds be?

 

[collinsmark]

the bounds on the left side is 0 to 45 correct? what would my left side bounds be?

Well, you're integrating over velocity, right? How about [itex] v_f [/itex] and [itex] v_0 [/itex]?

 

[erobz]

the bounds on the left side is 0 to 45 correct? what would my left side bounds be?

One is the speed you began with, the other is variable (final velocity) leave these all variable names though. You can integrate on the left from ##x_o## to ##x_f##, and as was mentioned ##v_o## to ##v_f## on the RHS.

 

[vbottas]

 

[vbottas]

View attachment 340082

i changed the left side back to xi and xf :)

 

[vbottas]

FTC time?

 

[collinsmark]

View attachment 340082

So far so good, except you left out the [itex] A [/itex] by accident somewhere along the line. But yes, that's the right idea.

Next step is algebra. Solve for [itex] v_f [/itex].

 

[erobz]

Now use log rules and solve for ##v_f##

 

[vbottas]

 

[erobz]

View attachment 340083

Keep simplifying. Log rules on the RHS to group those terms, then exponentiate each side.

 

[vbottas]

Keep simplifying

I can turn the rhs to ln(vf/vi) but that is the only simplificiation I see? Do you agree?

 

[erobz]

I can turn the rhs to ln(vf/vi) but that is the only simplificiation I see? Do you agree?

Next. exponentiation of each side.

 

[collinsmark]

I can turn the rhs to ln(vf/vi) but that is the only simplificiation I see? Do you agree?

Try taking both sides of the equation to the power of e. (I.e., exponentiate each side)

 

[vbottas]

Try taking both sides of the equation to the power of e. (I.e., exponentiate each side)

oh boy its been awhile since i've done this. the rhs ln disappears leaving just vf/vi. but for the left side, does all of the left side become the exponent for e?

 

[collinsmark]

oh boy its been awhile since i've done this. the rhs ln disappears leaving just vf/vi. but for the left side, does all of the left side become the exponent for e?

Yes.

 

[vbottas]

 

[vbottas]

seems like it's plug in time and thats it!

 

[collinsmark]

View attachment 340084

'Looks good to me.

All that's left to do is plug in the numbers.

 

[vbottas]

Wow. thank you so much. both of you, for your time and patience through this problem. I truly appreciate it. I could have hired a tutor and still wouldn't have recieved such great step by step help. Have a great rest of your day, yall deserve it! Take it easy!