##### Need help with setting up equation for physics

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Bus travels 280km south along a straight path with an average velocity of 88km/hr to the south. Bus stops for 24 min. Then travels 210km south with an average velocity of 75km/hr to the south.
How long does the total trip take?
What is the average velocity for the total trip?

Sep 10th, 2015

Thank you for the opportunity to help you with your question!

time taken to cover first 280 km

t1 =d1/v1= 280/88 = 3.18 hours =190.9 min

stopping time t2 =24 min

time taken to cover the last 210 km

t3 = 210/75= 2.8 hours = 168 mins

total time t = t1+t2+t3

t = 190.9+24+168 =382.9 mins

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 10th, 2015

average velocity

V = total displacement/total time

= (210+280)/6.38 =76.8 km/h

Sep 10th, 2015

Thank you once again! I think I have a better understanding of this.

Sep 10th, 2015

you are welcome leeza and thanks for giving me one more oportunity

Sep 10th, 2015

If it it possible, for future use, could you explain to me which equation I would have to use for what types of problems?

Sep 10th, 2015

you can post the problems and i would let you know

like here the equations necessary are

velocity = displacement /time

time = displacement/velocity

displacement= velocity  x time

post the other questions so that i can know if we need other equations

Sep 10th, 2015

Thank you!
1) What is the shortest possible time in which a bacerium could travel a distance of 8.4cm across a Petri dish at a constant speed of 3.5mm/s?

2)Child pushing shopping cart  speed of 1.5m/s. How long will it take this child to push the cart down an aisle with length of 9.3m?

3) An athlete swims from north end to the south end of a 50.0m pool in 20.0 sec and makes the return trip to the starting position in 22.0sec.
a.) what is the average velocity fot the first half of the swim?
b.) what is the average velocity for the second half of the swim?
c.) what is the average velocity for the roundtrip?

4) Two students walk in the same direction along a straight path, at a constant speed one at 0.90m/s and the other at 1.90m/s.
a.) assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780m away?
b.) How far would the students have to walk so that the faster student arrives 5.50min before the slower student?

Those are the only problems I have not done yet, I would really appreciate if you could just help me use the correct equation, and I then I could solve it.

Sep 10th, 2015

please request solutions for this questions these questions directly to me

only then i can help you with the solutions

Sep 11th, 2015

Okay.

Sep 11th, 2015

thanks leeza

Sep 11th, 2015

How am I supposed to send you the questions?

Sep 11th, 2015

you can post the question in standard questions and then send them to me

Sep 11th, 2015
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Sep 10th, 2015
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Sep 10th, 2015
Sep 20th, 2017
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