Kinematics - getting different answers when using different units

[shwatwat]

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?

Homework Equations

v²_fx = v²_ix + 2a_xΔ_x

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/s²)Δ_x
(-9m/s)Δ_x = 5625m²/s²
Δ_x = 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/s²)Δ_x
9m/s²Δ_x = 72900km²/hr²

and then an aside for unit conversion:

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

i plugged that into get

32.4km/hr²Δ_x = 72900km²/hr²
Δ_x = 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

 

[voko]

The conversion of the units in acceleration is way off.

 

[shwatwat]

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...

 

[voko]

3600 times 3600 over 1000 must be obviously greater than 1000.

 

[PhysicoRaj]

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)Δ_x
270²=116640*Δ_x
72900=116640*Δ_x
Δ_x=72900/116640
Δ_x=0.625km ( which is 625m)

 

[shwatwat]

thanks!