Calculating a Car Crash - Numberphile

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NOTE: The s in the SUVAT equations more precisely represents "displacement", but for the purposes of this video and to keep things simpler we used distance!

[henninghoefer]

In Germany, this (the quadratic behavior of braking distance in relation to speed) is required knowledge to get a driving license. The test questions even include how far you travel in the second you need to react.

[wvvwkx]

70 mph ≈ 113 km/h
100 mph ≈ 161 km/h
30 mph ≈ 48 km/h
71 mph ≈ 114 km/h

[matthewmontgomery]

I love the idea of the v-squared speedometer.

[danielrose]

Anyone asking for km/h or other speeds, there is a general rule for any unit.

For any speed x, traveling at sqrt(2)*x will double your kinetic energy.

The breaking distance for x to 0 is equal to the breaking distance from sqrt(2)*x to x.

Put in whatever numbers and units you like.

[nicosmind]

I was in a car crash, and if you can pick up from subtle clues, it didn't work out too well for me

[maxwvm]

It is not speed that causes injury, it is the sudden deceleration. For this reason I propose to make breaks illegal

[lodevijk]

That was an extremely elegant way of calculating that. And the idea about the speedometer is pretty nice as well.

[flymypg]

After 30+ years as an engineer, in two days I'm interviewing to become a high school STEM teacher. I'm going to use this simple (but not obvious) observation every chance I get.

[nicfink]

For those of you that are curious about the "hard" way to solve this:

Let v1 be the initial speed of the blue car and v2 the initial speed of the red car. The distance to the tree is D. Furthermore, the blue car comes to a halt after time t1, the red car will crash after time t2. The acceleration is a for both cars (see the footnote for why this is a reasonable assumption).

Then we have for the blue car:
(1) D=v1*t1+1/2*a*t1^2
(2) v1+a*t1=0 <=> a= -v1/t1
from (1) we replace a according to (2) and get
(3) D=v1*t1 - 1/2*v1*t1=1/2*v1t1 <=> t1 = 2*D/v1
replace t1 in (2) and get
(4) a = -v1^2/(2*D)

For the equations of the red car we have similarily:
(5) D=1/2*a*t2^2+v2*t2 <=> 1/2*a*t2^2+v2*t2 - D=0
This is a quadratic equation with the positive solution (we neglect the negative solution for obvious reasons):
(6) t2 = (-v2+SQRT(v2^2+2a*D))/a
The speed at time t2 with a constant acceleration a is given by
(7) v=v2+a*t2
we replace t2 in (7) and get
(8) v=v2+ (-v2+SQRT(v2^2+2a*D)) = SQRT(v2^2+2a*D)
we use the expression for a in (4) and get our final result:
v=SQRT(v2^2-v1^2)

As promised, i will try to reason why we can assume that a=constant for both cars.

For a moving object, the force of friction F_R is very nearly constant within a reasonably small range of velocities.
Newton tells us, that
(9) F_R = m*a
where m is the mass of the object (in our case, the mass of the cars). Since F_R is constant and the mass doesnt change either it follows, that the acceleration a is constant aswell.
Again, this isnt entirely true as the force of friction is dependant on the velocity. If anyone could provide some real world values on this relation, i'd love to know about it.

(Btw, assuming that F_R is constant we also get the conservation of energy as it was used in the video:
The energy loss through friction is:
W = (Integrate from starting point to end point) dx*F_R
which is the same for both cars (since the path is the same and F_R is the same for either car))

[boomerang]

This dude's hairstyle straight out of 1700's

[ZamanSiddiqui]

Here's the SUVAT solution:

Equations to use are:
v = u + at [1]
s = ut + 0.5at^2 [2]
s = 0.5(u + v)t [3]
v^2 = u^2 + 2as [4]
s = vt - 0.5at^2 [5]

Blue car: u = 70, v = 0, a?, t?, s?

Using s = 0.5(u + v)t:

s = 0.5(70+0)t = 35t

Using v = u + at:

0 = 70 + at, therefore a = - 70 / t

Given that a and s are the same for the red car:

Red car: u = 100, a = -70 / t, s = 35t, v?

v^2 = u^2 + 2as

v^2 = 100^2 + 2(-70/t)(35t)
v^2 = 5100
v = 71.4mph (3s.f)

So the red car hits the tree at 71.4mph.

[terryhollands]

Every new driver should watch this video.

[sankalpabanerjee]

The knight rider theme at the end was a nice touch!

[TheFrozenfish]

How about describing the speed in barleycorns per quarter-decade? That'll make things easier.

[thanasisgiannakopoulos]

That's a nice sixty symbols episode you have there

[brunomachado]

That music at the end! Took me back to my childhood! Knight rider...

[JanSanono]

This was so weird expecting an overtake on the left all the time

[Platanov]

I'm a game designer, and I have a saying that every part of making a game is somewhat tricky, except the User Interface, which is incredibly difficult. Ben's idea about speedometers showing speed in a non linear way is absolutely genius.

[feynstein]

That speedometer is lit. I want one. Hmm just have to get a car first.

[OutbackCatgirl]

I remember here in Australia they ran a bunch of speed awareness campaigns in the early 2000's on TV. One such campaign, which I believe you can find by searching "Slow-Mo TAC Anti-Speed TV Ad", explores a very similar scenario. Two identical cars, two identical reaction times, and an obstacle appearing in the road ahead. Just thought that might be interesting to share for anyone looking for another visual comparison.

There's a direct quote in the ad which stuck with me right through to today. "In the last five meters of braking, you wipe off half your speed."