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A Physics Question
In article >,
Scott en Aztl?n > wrote: > Imagine there is a 4-door sedan, such as a Honda Accord, parked on an > asphalt surface. In your estimation, how fast would a compact pickup > truck, such as a Ford Ranger, have to be travelling in order to strike > the Accord and move it SIDEWAYS (i.e. perpendicular to the direction > the wheels normally roll) by 15-18 inches? Imagine that the point of > contact betwen the two vehicles is the truck's bumper and the rear > quarter panel of the car. > > Follow-up question: how much damage would you expect to see on the two > vehicles? > > Thank you for your insight. This really isn't that hard a question. Assuming that the sedan's tires are the only ones providing the braking force then the question devolves to how fast the two vehicles can be moving at the instant after impact and be stopped in 18 inches. Since the combined masses of the vehicles will be roughly doubled, the maximum deceleration of the tires will be roughly halved. Final speed squared = initial speed squared + 2 acceleration * distance vf^2 = vi^2 + 2ad. a = 0.5g = 16ft/sec^2 d = 1.5ft vf = 0 Solve for vi: vi = (2ad)^(1/2) = 48^(1/2) = 6.9 ft/sec = 4.7 mph. But that is the speed of the two moving *after* collision. If they weigh roughly the same, then the truck will be moving roughly twice as fast before impact. 10 mph. -- Alan Baker Vancouver, British Columbia "If you raise the ceiling four feet, move the fireplace from that wall to that wall, you'll still only get the full stereophonic effect if you sit in the bottom of that cupboard." |
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#2
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A Physics Question
In article >,
Scott en Aztl?n > wrote: > Alan Baker > said in rec.autos.driving: > > >Assuming that the sedan's tires are the only ones providing the braking > >force then the question devolves to how fast the two vehicles can be > >moving at the instant after impact and be stopped in 18 inches. > > > >Since the combined masses of the vehicles will be roughly doubled, the > >maximum deceleration of the tires will be roughly halved. > > > >Final speed squared = initial speed squared + 2 acceleration * distance > > > >vf^2 = vi^2 + 2ad. > > > >a = 0.5g = 16ft/sec^2 > > Pardon me for asking, but how did you determine that the acceleration > is 0.5g? Are you making some assumptions about the coefficient of > sliding friction of the sedan's tires? I was somewhat generous in the figure. Rubber on asphalt can approach 1g when loaded with the entire mass it must accelerate, so when it must decelerate twice the mass from which it receiving downward load (i.e. when one vehicles tires are stopping both that vehicle and another of equal mass) the deceleration will be half that value. But since sliding friction is actually a little less than non-sliding friction, you may redo the math if you wish. -- Alan Baker Vancouver, British Columbia "If you raise the ceiling four feet, move the fireplace from that wall to that wall, you'll still only get the full stereophonic effect if you sit in the bottom of that cupboard." |
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