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Physics 14 Online
OpenStudy (anonymous):

How far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g

OpenStudy (anonymous):

is that 36 m/s or 36 miliseconds

OpenStudy (anonymous):

miliseconds

OpenStudy (anonymous):

you know what formula you have to use?

OpenStudy (anonymous):

nope im lost

OpenStudy (anonymous):

physics?

OpenStudy (anonymous):

you know a = -60g since you're slowing down. time is .36seconds finial velocity = 0m/s You can use the equation vfinal = vinitial + at 0 = vinitial -60g(.36seconds) vinitial = -60g(.36seconds) now you can use this kinematics equation xfinal - xinitial = vinitial t + 1/2at^2 xfinal - 0 = -60g(.36seconds) (.36) + 1/2 (-60g)(.36)^2

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