Question

A rocket ship travels 14,000 miles per hour. What is the rocket ship's velocity in m/s in correct scientific notation? (1 km = 0.62 mi)

A. 6.3 × 10^3 m/s

B. 0.627 × 10^3 m/s

C. 630 × 10^6 m/s

D. 6.3 × 10^5 m/s

Please explain?? THanks! :)

Answer 1

First convert to scientific notation:
\(14,000=1.4\times10^4\)
\(0.62=6.2\times10^{-1}\)

Now some dimensional analysis:
\[\frac{1.4\times10^4\text{ mi}}{1\text{ hr}}
\times
\frac{1\text{ km}}{1\text{ mi}}
\times
\frac{1\times10^3\text{ m}}{\text{ km}}
\times
\frac{1\text{ hr}}{6\times10^1\text{ min}}
\times
\frac{1\text{ min}}{6\times10^1\text{ sec}}\]

Answer 2

You should notice that the units will cancel to yield \(\dfrac{\text{m}}{\text{sec}}\). Now it's matter of computing the magnitude.

Answer 3

ohh okay :)

so from here, how do you compute the magnitude? :/

Answer 4

Small typo:\[\frac{1.4\times10^4\text{ mi}}{1\text{ hr}}
\times
\frac{1\text{ km}}{\color{red}{6.2\times10^{-1}}\text{ mi}}
\times
\frac{1\times10^3\text{ m}}{\text{ km}}
\times
\frac{1\text{ hr}}{6\times10^1\text{ min}}
\times
\frac{1\text{ min}}{6\times10^1\text{ sec}}\]

Answer 5

ah okay:)

Answer 6

\[\begin{align*}\frac{1.4\times10^4\times10^3}{6.2\times10^{-1}\times6\times10^1\times6\times10^1}&=\frac{1.4\times10^7}{223.2\times10^1}\\\\
&=\frac{1.4\times10^7}{(2.232\times10^{-2})\times10^1}\\\\
&=\frac{1.4\times10^7}{2.232\times10^{-1}}\\\\
&=\cdots\times10^{6}
\end{align*}\]
where \(\cdots=\dfrac{1.4}{2.232}\).

Answer 7

ohh isee :)

so 0.627 x 10^6 ?

:/

Answer 8

Right, or \(6.27\times10^5\). If you're accounting for significant digits, you should round up to two: \(6.3\times10^5\).

Answer 9

ahh okay :) thanks so much!! so basically for problems like this, you keep converting until you hit the scientific notation through simplification?

Answer 10

hmmm....

Answer 11

Right. It helps to write out each conversion factor like I did in the first comment. You don't have to convert to sci notation right away, it's just a way of easily seeing the powers of 10 reduce.

Answer 12

Did you double check your math Sith?

Answer 13

ahh okay:)

@e.mccormick ?

Answer 14

I did not :)

Answer 15

Just double-checked with wolfram, there's a mistake up there. Should be \(6.3\times10^3\) I believe...

Answer 16

ohhh okay was it just the calculation at the end? :/

Answer 17

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[\begin{align*}\frac{1.4\times10^4\times10^3}{6.2\times10^{-1}\times6\times10^1\times6\times10^1}&=\frac{1.4\times10^7}{223.2\times10^1}\\\\
&=\frac{1.4\times10^7}{\color{red}{(2.232\times10^{2})}\times10^1}\\\\
&=\frac{1.4\times10^7}{2.232\times10^{3}}\\\\
&=\cdots\times10^{4}
\end{align*}\]
\(\color{blue}{\text{End of Quote}}\)

Answer 18

ahh okay haha thank you so much!! :D

Answer 19

Thanks for catching it @e.mccormick

Answer 20

yes, thank you @e.mccormick !! :D

Answer 21

Way I started it was:

\(\dfrac{14000mi}{1h} \cdot \dfrac{1k}{.62mi} = \dfrac{22580.645k}{1h}\)

\(\dfrac{22580.645k}{1h} \cdot\dfrac{1h}{60min} = \dfrac{376.344k}{min}\)

When you finish with the min to sec and k to meter changes, you get the answer. You can see it is the same sort of thing.

As Sith pointed out, one of the keys is watching what cancels:

By putting miles on the bottom in the fist conversion, it cancels with the riginal miles.

\(\dfrac{14000\bbox[border:2px solid red]{mi}}{1h} \cdot \dfrac{1k}{.62\bbox[border:2px solid red]{mi}} = \dfrac{22580.645k}{1h}\)

Same sort of thing with hours, but this time it is on top to cancel what was on the bottom in the original.

\(\dfrac{22580.645k}{1\bbox[border:2px solid red]{h}} \cdot\dfrac{1\bbox[border:2px solid red]{h}}{60min} = \dfrac{376.344k}{min}\)

Answer 22

ahh okay :) so when i do these problems, i'm trying to work on canceling out until i reach the unit/simplification that i'm looking for?

Answer 23

Exactly!

Answer 24

awesome! I'll be trying to do some more practice problems so i can have this engrained in my brain hehe!! :)

Answer 25

If you are doing it in a calculator the whole way, the method I did works very well. When you are doing it by hand, the method where you trak and cancel \(\times 10^x\) works really well. Not a huge difference between them.

Answer 26

okay :) thank you!! i'll keep everything in mind (hopefully!!!) :D