Kinematics - getting different answers when using different units

  • Thread starter Thread starter shwatwat
  • Start date Start date
  • Tags Tags
    Kinematics Units
AI Thread Summary
The discussion revolves around calculating the minimum runway length required for a jetliner landing at 270 km/h, decelerating at 4.5 m/s². The correct conversion of speed from km/h to m/s is crucial, resulting in 75 m/s. A significant error occurs when converting the acceleration from m/s² to km/hr², where the correct conversion is -58320 km/hr², not 32.4 km/hr². This miscalculation leads to an incorrect runway length of 2250 km instead of the accurate 625 m. Proper unit conversion is essential for obtaining the correct results in kinematic equations.
shwatwat
Messages
3
Reaction score
0

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



v2fx = v2ix + 2axΔx

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 into 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:
Physics news on Phys.org
The conversion of the units in acceleration is way off.
 
voko said:
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...
 
3600 times 3600 over 1000 must be obviously greater than 1000.
 
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)
 
thanks!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top