How Do You Convert Acceleration from m/s^2 to km/h^2?

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The discussion centers on converting acceleration from meters per second squared (m/s²) to kilometers per hour squared (km/h²) in the context of calculating an airplane's velocity. The airplane accelerates at a constant rate of 3.00 m/s² for 30 seconds, resulting in a final velocity of 2.84 km/h. Participants emphasize using the kinematic equation Vf = Vi + at for direct calculations and suggest that unit conversions can be simplified by multiplying by "one" to maintain dimensional consistency. The final velocity calculation is confirmed to be correct based on the provided parameters.

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An airplane starts from rest and accelerates at a constant rate of 3.00 m/s^2 for 30.0s before leaving the ground. What is the airplanes velocity in km/h at the end of this interval?

Vi=0m/s
Vf=?
a=3m/s^2
Δt=30s


Not sure if I did it right or if I am also supposed to convert the acceleration to km/h before plugging into equation. If so, how do i convert 3m/s^2 to km/h^2?? Please help
 
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Just get the final velocity in m/s then convert to km/h. That would be the simplest thing to do.
 
Henrybar said:
An airplane starts from rest and accelerates at a constant rate of 3.00 m/s^2 for 30.0s before leaving the ground. What is the airplanes velocity in km/h at the end of this interval?

Vi=0m/s
Vf=?
a=3m/s^2
Δd=1350m
= 1.35km

Vf^2=Vi^2 + 2aΔd
Vf=2.84km/h

Not sure if I did it right or if I am also supposed to convert the acceleration to km/h before plugging into equation. If so, how do i convert 3m/s^2 to km/h^2?? Please help

You are given acceleration and time, not acceleration and distance. You need to use a different version of the kinematic equation -- one that relates velocity to acceleration and time...

And to do any units conversion, just multiply by "one". So if I wanted to convert inches to mm, for example, I'd multiply by 1 = (25.4mm/1 inch), cancel the "inch" units on the top and bottom of the fraction, and be left with the distance in mm. :smile:
 
Henrybar said:
Vf=2.84km/h

Not sure if I did it right

Check your answer using common sense. At that speed, if you were late for the check-in, you would be able to walk down the runway (not even run) and catch up the plane to get on the flight :smile:

Also, it's not "wrong" to find the distance and then the speed (if you do it right), but why not use the formula Vf = Vi + at?
 

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