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Kinematics - getting different answers when using different units

  • Thread starter shwatwat
  • Start date
3
0
1. Homework Statement

A jetliner touches down at 270 km/h. The plane then decelerates (i.e., undergoes acceleration directed opposite its velocity) at 4.5 m/s2. What’s the minimum runway length on which this aircraft can land?

2. Homework Equations

v2fx = v2ix + 2axΔx

3. The Attempt at a Solution

i got the solution doing this:

(270km/hr)(1000m/km)(hr/3600s) = 75m/s

0 = (75m/s)2 + 2(-4.5m/s2x
(-9m/s)Δx = 5625m2/s2
Δx = 625m

which is the correct answer.

however, i also tried to get the answer by converting everything to km/h, and i couldn't get the correct answer. i can't figure out why, and it's driving me crazy. here's what i tried:

0 = (270km/hr)2 + 2(-4.5 m/s2x
9m/s2Δx = 72900km2/hr2

and then an aside for unit conversion:

(9m/s2)(km/1000m)(3600s2/hr2) = 32.4km/hr2

i plugged that in to get

32.4km/hr2Δx = 72900km2/hr2
Δx = 2250 km, which is much, much more than the 625 m that i got above.

what am i doing wrong?!?

thanks for the help, i really appreciate it
 
Last edited:

Answers and Replies

6,049
390
The conversion of the units in acceleration is way off.
 
3
0
The conversion of the units in acceleration is way off.
how so?

this is probably a silly question but i've reworked this like four times and don't see what i'm doing wrong...
 
6,049
390
3600 times 3600 over 1000 must be obviously greater than 1000.
 
PhysicoRaj
Gold Member
473
33
You are wrong at converting -4.5 m/s2 to 32.4 km/hr2
it is -58320 km/hr2

4.5 m/s2=[4.5(10-3)]/(3600*3600)-1
=(4.5*3600*3600)/1000
=58320
Hence,
0=2702+2(-58320)Δx
2702=116640*Δx
72900=116640*Δx
Δx=72900/116640
Δx=0.625km ( which is 625m)
 
3
0
thanks!
 

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