Calculating Average Acceleration of a 747 Airplane

[dg_5021]

A 747 airliner reaches its takeoff speed of 173 mi/h in 35.2s. What is the magnitude of its average acceleration?

(77.3341 m/s)/(35.2 s) = 2.20 m/s^2

I am wondering did I do it correctly?

 

[quasar987]

Except for the transformation from miles to meter, you sure did! And by "except", I don't mean that your transformation is wrong, I only mean that I don't know how to do the transformation, so I can't verify it. But the important part is that you used the right concept, that is,

$$a_{av} = \frac{\Delta v}{\Delta t}$$

and did not forget yo convert the units of time so they match.

 

[plusaf]

what means "correctly"?

A 747 airliner reaches its takeoff speed of 173 mi/h in 35.2s. What is the magnitude of its average acceleration?

(77.3341 m/s)/(35.2 s) = 2.20 m/s^2

I am wondering did I do it correctly?

please clarify what you mean by that?
getting the right answer and "doing it correctly" are very different things. did your teacher/instructor/whatever want to see all of the steps involved? if so, your answer may be right but the "doing it" is invisible and can't be judged!

the "process" might be something like this:

((173 mi/hr)*(# ft/mile)*(#m/ft)*(1 hour/60min)*(1min/60 s))/35.2 s.

if i got all of the conversions in there and all of the units cancel right, the answer will be right for acceleration = delta v / delta t.

did that make sense?
cheers!
+af

 

[dextercioby]

$$1Mph=\frac{1609.344}{3600}ms^{-1}$$

Daniel.

 

[plusaf]

ok, i'll bite...

A 747 airliner reaches its takeoff speed of 173 mi/h in 35.2s. What is the magnitude of its average acceleration?

(77.3341 m/s)/(35.2 s) = 2.20 m/s^2

I am wondering did I do it correctly?

if the problem is stated in miles/hour and seconds for acceleration, what was the purpose of converting to meters/second? was that required by the problem, or are all problems' answers required to me in the metric system?

 

[dextercioby]

Of course not...But in this case it would be rather awkward to express the acceleration in $M hr^{-2}$,don't u think...?

Daniel.