1018
  • 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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
1018
  • 1018
my answer for (a) is (79+105)/2 = 92 and (d)= 0. are these correct?
anonymous
  • anonymous
Yes.
1018
  • 1018
thanks. but how do i solve for b and c?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
b) is done exactly the same way as a). What do you think?
1018
  • 1018
yeah, that's what i thought. that it would be tha same but it was wrong when i tried it.
1018
  • 1018
i also found a similar question with different values, and they a and b werent the same either
anonymous
  • anonymous
Can't understand why. Half the trip is at 79 and the other half is at 105.
1018
  • 1018
i know! me neither
1018
  • 1018
and c would be just a+b / 2 it was also wrong
anonymous
  • 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
  • 1018
OOOH I didnt see that hahaha
1018
  • 1018
wait ill try haha. can you answer it too? just to see if im right
anonymous
  • anonymous
Sure. Go ahead.
anonymous
  • 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
  • 1018
i got 90? is that correct? @Nishant_Garg
anonymous
  • anonymous
that's a good enough answer, it will not come as a round figure, I'm getting around 89.19
1018
  • 1018
is it ok? originally i got 90.12 something. was that wrong or that's acceptable already?
1018
  • 1018
oh nevermind, 90 was correct. haha. thanks!
anonymous
  • anonymous
for 3rd I assume you take the average of the forward and back trip's speed's
1018
  • 1018
yes that's what i did, and it's correct
anonymous
  • anonymous
actually yeah, I made a slight mistake 90.16 is the correct answer so approximately 90
anonymous
  • anonymous
for the 2nd part
anonymous
  • anonymous
so for 3rd it should average of 90 and 92, so 91
1018
  • 1018
yes, all answers were right. thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.