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- 1018

(physics) You drive on Interstate 10 from San Antonio to Houston, half the time at 79 km/h and the other half at 105 km/h. On the way back you travel half the distance at 79 km/h and the other half at 105 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip?

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- 1018

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- 1018

my answer for (a) is (79+105)/2 = 92 and (d)= 0. are these correct?

- anonymous

Yes.

- 1018

thanks. but how do i solve for b and c?

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- anonymous

b) is done exactly the same way as a). What do you think?

- 1018

yeah, that's what i thought. that it would be tha same but it was wrong when i tried it.

- 1018

i also found a similar question with different values, and they a and b werent the same either

- anonymous

Can't understand why. Half the trip is at 79 and the other half is at 105.

- 1018

i know! me neither

- 1018

and c would be just a+b / 2 it was also wrong

- anonymous

Oh, I get it now. The difference is, for the return trip, half the DISTANCE is at 79 and the other at 105. Gotta read carefully.

- 1018

OOOH I didnt see that hahaha

- 1018

wait ill try haha. can you answer it too? just to see if im right

- anonymous

Sure. Go ahead.

- anonymous

for 2nd consider the the distance between the 2 places be
\[s=2x\]
Assume time taken
\[t_{1}\]
for first half of the distance and time
\[t_{2}\]
for the 2nd half
\[v_{av}=\frac{s}{t}=\frac{s}{t_{1}+t_{2}}\]
Let us find t1 and t2,
\[t_{1}=\frac{s/2}{79}=\frac{x}{79}\]
similarly
\[t_{2}=\frac{x}{105}\]\[v_{av}=\frac{2x}{\frac{x}{79}+\frac{x}{105}}\]
On simplification the x will be cancelled, alternatively you could do this with just the s and without introducing the x

- 1018

i got 90? is that correct? @Nishant_Garg

- anonymous

that's a good enough answer, it will not come as a round figure, I'm getting around 89.19

- 1018

is it ok? originally i got 90.12 something. was that wrong or that's acceptable already?

- 1018

oh nevermind, 90 was correct. haha. thanks!

- anonymous

for 3rd I assume you take the average of the forward and back trip's speed's

- 1018

yes that's what i did, and it's correct

- anonymous

actually yeah, I made a slight mistake 90.16 is the correct answer
so approximately 90

- anonymous

for the 2nd part

- anonymous

so for 3rd it should average of 90 and 92, so 91

- 1018

yes, all answers were right. thanks!

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