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rjandmj82 rjandmj82
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12 years ago
A speeder passes a parked police car at 34.9
m/s. Instantaneously, the police car starts
from rest with a uniform acceleration of 2.42
m/s2.

a) How much time pases before the speeder
is overtaken by the police car?

b) How far does the speeder get before being
overtaken by the police car?

help please!
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Answer accepted by topic starter
lenorefrancislenorefrancis
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#1
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12 years ago
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#2
12 years ago
x=(1/2)at^2 + Vo + Xo on both cars
set the x formulas equal to each other. and use the quadratic formula to solve for the time.  then multiply the time by the speed and u get the distance



A= 28.8s
B=1005.12 m (w/o considering sig figs)
B= 1.01x10^3  m (with significant figures)
wrote...
#3
12 years ago
yeah i worked it out. Jon has the correct answer.
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