The discussion focuses on calculating the final velocity of a car moving with an initial speed of 30.63 m/s over a distance of 45 meters, while experiencing a decreasing drag force. The drag force is defined by the equation F_d = (1/2) * C_d * A * ρ * v^2, where C_d is the drag coefficient (0.20), A is the cross-sectional area (8.71 m²), and ρ is the air density (1.3 kg/m³). Participants emphasize the importance of applying Newton's second law and converting the differential equation to integrate over distance rather than time, ultimately leading to the calculation of the final velocity.
PREREQUISITESPhysics students, automotive engineers, and anyone interested in understanding the dynamics of vehicles under the influence of drag forces.
Yes, that's right.vbottas said:
collinsmark said:
vbottas said:
should I plug in known values to the left side?collinsmark said:
Perfect.vbottas said:
Now integrate both sides of the equation. (This will involve two integrals, one on each side of the equation.)Always try to work in variables till the end. Calculate as last step.vbottas said:
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 said:
The constants come outside the integral.vbottas said:
the bounds on the left side is 0 to 45 correct? what would my right side bounds be?collinsmark said:Perfect.
Well, you're integrating over velocity, right? How about v_f and v_0?vbottas said:
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 said:
i changed the left side back to xi and xf :)vbottas said:
So far so good, except you left out the A by accident somewhere along the line. But yes, that's the right idea.vbottas said:
Keep simplifying. Log rules on the RHS to group those terms, then exponentiate each side.vbottas said:
I can turn the rhs to ln(vf/vi) but that is the only simplificiation I see? Do you agree?erobz said:
Next. exponentiation of each side.vbottas said:
Try taking both sides of the equation to the power of e. (I.e., exponentiate each side)vbottas said:
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 said:
Yes.vbottas said:
'Looks good to me.vbottas said: