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1018
 one year ago
(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?
1018
 one year ago
(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
 one year ago
Best ResponseYou've already chosen the best response.1my answer for (a) is (79+105)/2 = 92 and (d)= 0. are these correct?

1018
 one year ago
Best ResponseYou've already chosen the best response.1thanks. but how do i solve for b and c?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0b) is done exactly the same way as a). What do you think?

1018
 one year ago
Best ResponseYou've already chosen the best response.1yeah, that's what i thought. that it would be tha same but it was wrong when i tried it.

1018
 one year ago
Best ResponseYou've already chosen the best response.1i also found a similar question with different values, and they a and b werent the same either

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can't understand why. Half the trip is at 79 and the other half is at 105.

1018
 one year ago
Best ResponseYou've already chosen the best response.1and c would be just a+b / 2 it was also wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, 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
 one year ago
Best ResponseYou've already chosen the best response.1OOOH I didnt see that hahaha

1018
 one year ago
Best ResponseYou've already chosen the best response.1wait ill try haha. can you answer it too? just to see if im right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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
 one year ago
Best ResponseYou've already chosen the best response.1i got 90? is that correct? @Nishant_Garg

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's a good enough answer, it will not come as a round figure, I'm getting around 89.19

1018
 one year ago
Best ResponseYou've already chosen the best response.1is it ok? originally i got 90.12 something. was that wrong or that's acceptable already?

1018
 one year ago
Best ResponseYou've already chosen the best response.1oh nevermind, 90 was correct. haha. thanks!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 3rd I assume you take the average of the forward and back trip's speed's

1018
 one year ago
Best ResponseYou've already chosen the best response.1yes that's what i did, and it's correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually yeah, I made a slight mistake 90.16 is the correct answer so approximately 90

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for 3rd it should average of 90 and 92, so 91

1018
 one year ago
Best ResponseYou've already chosen the best response.1yes, all answers were right. thanks!
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