Chrisupra said:
Hey guys,
I've been looking around online and only find formulae for car passing through a distance with an initial velocity but nothing to calculate the speed of an object starting from a stance and accelerating to complete a distance.
I have a car that has done a 1/4 mile (375m) and completes it in 7.671 seconds. I calculated the acceleration at 12.74m/s. Can anyone help me figure out the speed across the line?
I've tried .25/ (7.671/3600) and I get 117mph which is wrong..
Firstly: A 1/4 mile is 1,320 feet, which is “402.336 meters” (not 375 meters).
Secondly: A car traversing a distance with a steady initial velocity (merely for clarity’s sake since you stated that you found this equation) is computed as follows:
v = d / t
Naturally, this equation is useless for accelerated motion.
Thirdly: A car that accelerates from its state of rest to its highest velocity (as in the 1/4 mile) will not experience “uniform acceleration” over the 1/4 mile distance, as air resistance will exponentially increase per increases in velocity thereby significantly reducing the car’s rate of acceleration as velocities increase.
Even if you attempted to employ a drag coefficient for air resistance as the car’s velocity increases, headwinds and tailwinds further alter its potential top speed and gearing changes via up-shifts to taller gearing further increase the time required for the engine to increase its RPM to achieve its power-band therefore, the car’s top speed cannot accurately be calculated via simplistic kinematics equations, which deal with “uniform acceleration”. These equations only work in physics scenarios where the acceleration of the car is accepted as exhibiting a “uniform rate of acceleration”.
Typically, two beams positioned very close together whereby one is positioned slightly before the finish line, the other at an equal distance after the finish line, can be used to calculate the top speed by dividing the difference of the distance that separates the upper and lower beams by the difference of the two trip times. This yields a fairly miniscule acceptable error in 1/4 mile speed.
v = (d2 – d1) / (t2 – t1)
For instance, if the beams were positioned 1 meter before and 1 meter after the finish line (a difference in distance between the two beams of 2 meters) and the difference between the two trip times was detected as being .024717 seconds, it would indicate that a speed of 181 MPH had been achieved across the finish line:
v = d / t
2 meters / .024717 seconds = 80.916 m/s (181 MPH)
Since each of the trip times are sampled at equal distances before and after the finish line, they are times per velocities that are slightly lesser and slightly greater than would be achieved at the finish line. Since the lesser velocity causes the first distance (beam 1 to finish line) to be traversed at a slightly slower velocity (hence a slightly longer time), but the greater velocity (achieved between the finish line and beam 2) causes the second distance to be traversed at a slightly greater velocity (hence a slightly shorter time), the difference between these two times effectively average themselves to reveal the velocity that will be achieved as the car crosses the finish line.