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Car Physics porblem (was 3.2% of crashes caused by speeding - official)

Car Physics porblem (was 3.2% of crashes caused by speeding - official) - Physics Forum

Car Physics porblem (was 3.2% of crashes caused by speeding - official) - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


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  #1  
Old 11-15-2003, 12:50 PM
Oliver Keating
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Default Car Physics porblem (was 3.2% of crashes caused by speeding - official)




"scott" <[Only registered users see links. ]> wrote in message
news:bp268o$1bmc$[Only registered users see links. ]...
something
of

Cross posted to


KE = 0.5mv^2

Lets drop the units for conveniece.

Energy diff between cars doing 70 and 80, = 80^2 - 70^2 = 1500
ditto 10 and 20 = 20^2 - 10^2 = 300

It is the v^2 term which causes this discrepency. At the higher speeds there
is more energy to absorb even though the relative velocities are the same.

if
the

But it isn't, becuase if the car being hit is already travelling at 70, it
is already experiencing a great deal of frictional resistance, so any force
to change its speed has to overcome that first. Also the wheels have to gain
a lot more energy as rotational kinetic energy goes like v^2 too.

down!

No

the
might
over

During a head on collision the road plays a part because both vehicles have
a natural tendency to come to a stationary stop. I.e. frictional resistance
is helping to reduce the impact forces. This is not the case when a moving
car hits a stationary one, because frictional resistance of the stationary
car works against the impact force.

that


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  #2  
Old 11-15-2003, 02:51 PM
scott
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Default Car Physics porblem (was 3.2% of crashes caused by speeding - official)

"Oliver Keating" <[Only registered users see links. ].ac.uk> wrote in message
news:bp57cm$261s$[Only registered users see links. ]...

greater
a

there

But you can't do that, to calculate the KE you *need* to use the relative
velocity before you square. Otherwises as Paul said, you get an absurd
situation where someone at the equator has more KE difference than someone
at the poles !!!

by
force

No, assuming a car is travelling at a *constant* 70, the engine is already
overcoming the resistive forces. Any extra force will produce an
acceleration of exactly the same magnitude no matter how fast the car is
already going.


You are confused between forces/torques and energy. You're right in a way,
the *same* force/torque but at a higher speed will generate more energy (as
power = force * speed), but the *force* is the same.

have

This doesn't really have any effect during the short time of impact - tyre
friction forces are tiny compared to the impact forces. Anyway, tyre
friction forces are very similar over all speed ranges, including
stationary!

Scott


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  #3  
Old 11-15-2003, 04:40 PM
Phil Holman
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Default Car Physics porblem (was 3.2% of crashes caused by speeding - official)


"scott" <[Only registered users see links. ]> wrote in message
news:bp5ejv$9nh$[Only registered users see links. ]...
message
message
case of
speeds
same.
relative
absurd
someone

Frame of reference is important here. If this is a stationary observer
in space then the situation is not absurd. There actually is a KE
difference between the poles and the equator. However, the appropriate
frame of reference in this case (colliding cars) is to consider one of
the cars as stationary. Then the actual energy dissipation is correct
relative to the actual bodies involved and not to some arbitrary frame
of reference where energies are observed as being different.

will be
decreases
70, it
any
already
is
way,
energy (as

The force to accelerate a body through a 5mph increase in velocity is
the same, independent of original velocity.
However the energy requirement will be higher for greater initial
velocities. Energy is force times distance moved so at higher velocities
the force is applied through a much greater distance.

Phil Holman


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  #4  
Old 11-16-2003, 03:40 AM
Edward Green
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Default Car Physics porblem (was 3.2% of crashes caused by speeding - official)

"Phil Holman" <[Only registered users see links. ]> wrote in message news:<MFstb.1292$[Only registered users see links. ].pas.earth link.net>...


Actual energy dissipation will be a frame invariant.
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  #5  
Old 11-16-2003, 04:50 AM
Phil Holman
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Default Car Physics porblem (was 3.2% of crashes caused by speeding - official)


"Edward Green" <[Only registered users see links. ]> wrote in message
news:2a0cceff.0311151940.4501c428@posting.google.c om...
news:<MFstb.1292$[Only registered users see links. ].pas.earth link.net>...
observer
appropriate
of
correct
frame

Yes 'frame-invariant' meaning if E is the energy of a moving object in
my frame of reference measured by me, then any frame of reference you
are in will always answer that I measured the energy as E. This does not
mean that the quantity E is observer independent.

Phil Holman


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  #6  
Old 11-17-2003, 04:24 AM
Edward Green
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Default Car Physics porblem (was 3.2% of crashes caused by speeding - official)

"Phil Holman" <[Only registered users see links. ]> wrote in message news:<ZlDtb.1958$[Only registered users see links. ].pas.earthl ink.net>...

<speaking of collisions>


Yes, but that's not what I meant.

I meant that for any interacting system of bodies, the total change of
the kinetic energy for any process is a frame invariant -- a genuine
one, the same for all observers as measured in their own proper
coordinates, not the dilute second order kind you are talking about.

Here's a short proof.

Assuming the masses are constant ...

KE_tot = sum[(1/2)m_i(v_i)^2]

d/dt {KE_tot} = sum[m_i v_i (d/dt)v_i] = sum[v_i(d/dt)p_i]

Ok, that's in one frame. Now make a galilean transformation to
another frame: we add some v' -- the same v' -- to all v_i's. So the
time derivative of KE_tot changes by a term of form v' sum[(d/dt)p_i].

But, the object inside the last sum is just f_i, and sum[f_i] = 0 by
Newton's third law so long as all forces are interaction forces
between the bodies. So as long as we consider a closed system (don't
have any unpaired forces) the time rate of change of KE is a true
frame invariant.

OTOH, perhaps you were referring to the energy dissipation of a
particular vehicle, which is not an invariant, and I shouldn't have
butt in. When we change frames the energy accounting moves around,
even though the net change is invariant.
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