Calculate Distance of E. coli Bacteria w/ Acceleration of 151

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The discussion revolves around calculating the distance an E. coli bacterium needs to travel to reach a speed of 12, given an acceleration of 151. The relevant equation used is v^2 = v0^2 + 2ax, where the initial speed (v0) is zero. The calculation yields a distance of 0.48, which is confirmed as correct by other participants. The thread also indicates a request for further assistance on another problem. Overall, the focus is on applying physics equations to solve a specific biological scenario.
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Homework Statement


Approximately 0.1 of the bacteria in an adult human's intestines are Escherichia coli. These bacteria have been observed to move with speeds up to 15 and maximum accelerations of 166 . Suppose an E. coli bacterium in your intestines starts at rest and accelerates at 151 .
How much distance is required for the bacterium to reach a speed of 12 ?

Homework Equations


d=vt
t=v/a
v^2=v0^2+2ax


The Attempt at a Solution


i used v^2=v0^2+2ax
so v0 is 0 (i thiink)
12^2=2(151)x
x=.48
is this correct??
 
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Depending on if your units match, it looks OK.
 
thanks it was correct...could you help me with one more problem?
 

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