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

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SUMMARY

The discussion focuses on calculating the distance an Escherichia coli bacterium must travel to reach a speed of 12 m/s, given an acceleration of 151 m/s². The participant correctly applied the kinematic equation v² = v₀² + 2ax, where the initial velocity (v₀) is 0. The calculated distance (x) required for the bacterium to reach the desired speed is 0.48 meters. The solution was confirmed as correct by another forum member.

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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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