Can Car B Make it Past the X in Time to Avoid Disaster?

[danielatha4]

Car A is dropped at 4000 ft where directly below is an x on the road. At the instant Car A is dropped Car B is exactly 4000 ft from the x driving down the road at a constant velocity of 142 mph. Does Car B make it past the x in time to avoid disaster?

It's such a simple kinematics problem that it's beautiful but unfortunately you have to take into consideration wind resistance Can anyone show an accurate application of wind resistance in this problem? From my understanding the mass of the car is 1600 kg and it dropped staying flat landing virtually on all 4 tires at the same time.

For the heck of it I'll do the kinematics neglecting wind resistance:

time Car B takes to get to the x

x=vt
1219.2m = 63.5 m/s * t
t = 19.2 seconds

Does Car A hit the road is less time or more time than 19.2 seconds?

x=v_ot+1/2at²
x = 0 + 4.9 * 19.2²
x = 1806.336m

therefore, less time

time it takes Car A to hit the road

x = 1/2at²
1219.2m = 4.9 * t²
t = 15.77 seconds

Seems like air resistance and terminal velocity make all the difference here

 

[russ_watters]

142 mph would probably be pretty close to the terminal velocity of a car (maybe even above), so I expect the driving car would get there first.

 

[mgb_phys]

Terminal velocity is sqrt( 2mg / density A Cd )

BMW 3 series = 4.4m * 1.7m 1465kg
Assuming that on four wheels Cd = 1
And air is density = 1.2kg/m^3

terminal V = sqrt ( (2 * 1465 * 9.8) / (1.2 * 4.4 * 1.7 * 1)) = 56.5 m/s = 126mph

 

[danielatha4]

Does anyone know how to setup the differential equation needed to map out the total time the Car takes to fall 4000 ft (1219.2m)?

Force due to drag is
F_d=1/2[tex]\rho[/tex]v²AC_d

Where v seems to be the changing variable here. Therefore, force due to drag is a function of velocity in this case?

 

[diazona]

142 mph would probably be pretty close to the terminal velocity of a car (maybe even above), so I expect the driving car would get there first.

In the episode in question, if I remember correctly, they could only get their car up to 120-ish mph, so they adjusted the horizontal distance to the target accordingly... it wasn't much of a physics experiment. But still cool I think the falling car hit first by a second or two.

 

[diazona]

Does anyone know how to setup the differential equation needed to map out the total time the Car takes to fall 4000 ft (1219.2m)?

Force due to drag is
F_d=1/2[tex]\rho[/tex]v²AC_d

Where v seems to be the changing variable here. Therefore, force due to drag is a function of velocity in this case?

Sure, just use [itex]F = ma[/itex], remembering that [itex]a = \ddot{y}[/itex] and [itex]v = \dot{y}[/itex] (dot stands for the time derivative, of course).

[tex]\frac{1}{2}\rho A C_d \dot{y}^2 - mg = m\ddot{y}[/tex]

oooh, wait - is this a homework assignment?

 

[russ_watters]

Terminal velocity is sqrt( 2mg / density A Cd )

BMW 3 series = 4.4m * 1.7m 1465kg
Assuming that on four wheels Cd = 1
And air is density = 1.2kg/m^3

terminal V = sqrt ( (2 * 1465 * 9.8) / (1.2 * 4.4 * 1.7 * 1)) = 56.5 m/s = 126mph

...and that's if it stays stable and nose-down...

When they meausred the terminal velocity of a falling (not spin stabilized) bullet, it fell sideways and was roughly stable in that configuration.

 

[russ_watters]

In the episode in question, if I remember correctly, they could only get their car up to 120-ish mph, so they adjusted the horizontal distance to the target accordingly... it wasn't much of a physics experiment. But still cool I think the falling car hit first by a second or two.

[googles] 105mph, and with the falling car hitting a full 2 seconds before (308 feet at 105 mph), I wonder what they scaled their distances to?

...and btw, the falling car fell sideways, though oscillating somewhat. Here's the video: http://rippol.com/watch/mythbusters/car-drop-minimyth/?t=3025&v=2169551

 

[danielatha4]

They scaled the distance to 2950 feet I believe.

I'm still trying to straighten up that differential equation to figure out an accurate time.

 

[mgb_phys]

You can do it easily with a spreadsheet, I get 90% of terminal roughly after 12.5 secs and a distance of 1220m in 25.5 secs
In that time a car doing 140mph would cover about 1450m

 

[danielatha4]

I'll take your word for it. I was going to try to use the math, not a spreadsheet, but oh well.

if it takes 25.5 seconds to fall then the driving car will pass the x 9.7 seconds before impact. Hmm

And no diazona, this wasn't a homework assignment. Just physics for fun, ha

 

[danielatha4]

So the car falling hit the x 2 seconds before the car driving reached it. That means it took 17.2 seconds.

How can you calculate 17.2 seconds from the differential equation?

 

[diazona]

As long as you're not trying to leech answers off us ...

[tex]\frac{1}{2}\rho A C_d \dot{y}^2 - mg = m\ddot{y}[/tex]

That's actually an integrable equation,
[tex]\frac{1}{2}\rho A C_d \dot{y}^2 - mg = m\frac{\mathrm{d}\dot{y}}{\mathrm{d}t}[/tex]

[tex]\int_0^t\mathrm{d}t = \frac{1}{g}\int_0^t\frac{\mathrm{d}\dot{y}}{\frac{\rho AC_d}{2mg} \dot{y}^2 - 1}[/tex]
You can do the integral on the right using the substitution [itex]u = \dot{y}\sqrt{(\rho A C_d/2mg)}[/tex], getting
[tex]t = \sqrt{\frac{m}{2\rho gAC_d}}[\ln(1 - u) - \ln(1 + u)]_0^t[/tex]
and if you solve that for [itex]\dot{y}[/itex], taking into account that [itex]\dot{y}(0)=0[/itex], you get
[tex]\dot{y} = -\sqrt{\frac{2mg}{\rho A C_d}}\tanh\biggl(\sqrt{\frac{\rho gAC_d}{2m}}t\biggr)[/tex]
Then you integrate it again to get
[tex]y = -\frac{2m}{\rho A C_d}\ln\cosh\biggl(\sqrt{\frac{\rho gAC_d}{2m}}t\biggr)[/tex]
Using mgb_phys's values with y=-4000ft, I get 25.6 seconds, same as the spreadsheet analysis. (No surprise there) Though these calculations are never really exact - there are a lot of things that could have been different in the actual experiment, like a higher mass, lower air density, or a lower drag coefficient. Any of those would reduce the fall time.