A car traveling at 30 m/s undergoes a constant negative acceleration of magnitude 2.00 m/s^2 when the brakes are applied. How many revolutions does each tire make before the car comes to a stop assuming the car does not skid and the tires have a radii(plural form of radius) of .300 meters?
2006-12-10 16:47:46 · 5 answers · asked by elguapo_marco_2008@sbcglobal.net 3 in Science & Mathematics ➔ Physics
the kinetic energy due to dpeed will become zero at some point.
we have to find the distance travelled till then. in order to do so we apply the theorem of change in kinetic enerfy and we end up to .
1/2 m * U^2 = F*d
but F = m*a (due to the deceleration)
therefore we end up to
d = U^2 / 2a => d = 30*30/4 = 225m
now Cir = 2 * 3.14 * r = 2*3.14*.3 = 1.885m
==>> 225m / 1.885 = 119.36 revolutions
2006-12-10 18:12:25 · answer #1 · answered by Emmanuel P 3 · 0⤊ 0⤋
Zidane is correct except for the mistakes committed in the calculation.
d = V^2/ (2a)
= 30 x 30 / (2x2) = 225m.
Circumference of the wheel = 2 x pi x radius
= 2 x pi x 0.3 = 1.885 m
Number of revolutions = distance / circumference of the wheel.
= 225m/1.885 m = 119 .4 revolutions.
2006-12-11 02:56:39 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
If the tire radius is 0.3m, then the circumference is 2*pi*r = 1.89 m
2006-12-11 01:10:29 · answer #3 · answered by arbiter007 6 · 0⤊ 0⤋
Vf=0m/s
Vi=30m/s
a=-2.00
d=?
we will use the kinematic equation to find distance:
Vf^2=Vi^2+2ad
Vf^2-Vi^2/2a=d
d=(0m/s)-(-30m/s)^2/2(2.00m/s^2)
d=225m
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since radii of circle is 0.300m, then diameter is 2r=d=0.600m
in order to go one revolution 0.600m must be travelled threfore to find number of diameter traveled divide distance by diameter=0.600m
225m/0.6m=375 circular revolutions
2006-12-11 00:50:52 · answer #4 · answered by Zidane 3 · 0⤊ 1⤋
do your homeowork yourself
2006-12-11 02:50:48 · answer #5 · answered by Neo 2 · 0⤊ 0⤋