According to recent typical test data, a Ford Focus travels 0.260 mi in 18.8 s ,

Question

According to recent typical test data, a Ford Focus travels 0.260 mi in 18.8 s , starting from rest. The same car, when braking from 61.5 mi/h on dry pavement, stops in 141 ft . Assume constant acceleration in each part of its motion, but not necessarily the same acceleration when slowing down as when speeding up. Find the magnitude of this car's acceleration while braking. Find the magnitude of this car's acceleration while speeding up. If its acceleration is constant while speeding up, how fast (in mi/h) will the car be traveling after 0.260 mi of acceleration? How long does it take the car to stop while braking from 61.5 mi/h ?

Explanation / Answer

while braking

initial velocity vox = 61.5 mi/h

distance x = 141 ft = 0.0267 mi

final velocity vx = 0

from equation of motion

vx^2 - vox^2 = 2*a*x

0^2 - 61.5^2 = 2*a*0.0267

acceleration a = 70828.65 mi/h^2    = 28.86 ft/s^2 <<<<-------answer

+++++++++

initial velocity vox = 0mi/h

distance x = 141 ft = 0.26 mi

time = t = 18.8 s

from equation of motion

x = vox*t +0.5*ax*t^2

0.26 = 0 + (0.5*ax*18.8^2)

acceleration ax = 0.0015 mi/s^2 = 7.92 ft/s^2 <<<<-------answer

+++++++++++++

v = vox + ax*t

v = 0 + (0.0015*18.8) = 0.0282 mi/s = 101.52 mi/h <<<<-------answer

++++++++++++++

v = vox + a*t

0 = 61.5 - (70828.65*t)

t = 0.000868 h = 3.126 s <<<<<-------answer

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