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Motion and force problems

  1. Mar 9, 2007 #1
    1. The problem statement, all variables and given/known data

    Vic was watching a car race on TV. At the instant the flag was lowered to start the race, the picture on TV screen goes out due to surge in the power. When the picture come back on TV, the timer on score board reads 75 s. At this point Vic observes that leading car was on opposite side of the racing track (opposite side to that where racing was started). The racing track is oval in shape and 6 Km in length.
    a) Determine leading cars average velocity during the time when TV was without picture?
    b) What are two possible distances leading car travelled when TV was without picture?
    c) Given the record for fastest racing car is 450 Km/hr, which is most likely distance-leading car has travelled when TV was without picture? ☺☺
    d) Based on your calculation in (c), calculate leading car average speed when TV was without picture?


    2. Relevant equations

    v = d/t


    3. The attempt at a solution

    I thought maybe that half way around the track would be 3 km, but then i need the displacemet? so would that be zero?

    o/75= 0 so that doesn't make sence

    3/75= 0.04 m/s

    004 m/s was the speed of the leading car
    But now i don't know ANYTHING
    I need some guidance, once i get on the right track i'll probably be able to finish the question.
     
  2. jcsd
  3. Mar 9, 2007 #2

    Dick

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    For a) you can't actually calculate the displacement since you don't know the width of the track (unless it's supposed be 0 - figure 8 track?). Anyway, I would just continue on with the rest of the questions. You are doing ok computing average speed except that the track is 6 km long, not 6 m.
     
  4. Mar 10, 2007 #3
    Thankyou, but what i am really confused about is b c and d, how am i sopposed to find two difernt distances?
     
  5. Mar 10, 2007 #4

    Dick

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    How many circuits of the track did the car do while the TV was off? Can you be sure it was on the first one?
     
  6. Mar 10, 2007 #5
    I never thought of that, is there any way of knowing?
     
  7. Mar 10, 2007 #6

    Dick

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    Uh, that's what the question is all about. You'll find out. What are the smallest two possibilities for distance? That's the answer to b). Now continue.
     
  8. Mar 10, 2007 #7
    Oh, so c is asking if it was going 450 km/h which distance from B would it be?

    So you have to use

    distance = (veloctiy) (time)
    (450) ( 75 s )
    = 33750

    Wait, should i be conveerting something? because i am using a mix of seconds and minutes?
     
  9. Mar 10, 2007 #8

    Dick

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    Yes, you should be converting. Your answer would be correct if the speed were 450 km/s. How many seconds in an hour? Please put units on things.
     
  10. Mar 10, 2007 #9
    3600 seconds in 1 hour

    75s/3600= 0.021 hours

    distance = (veloctiy) (time)
    (450km/h) ( 0.021h )
    = 9.4km/h
     
  11. Mar 10, 2007 #10

    Dick

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    Better. But (450km/h)(0.021 h)=9.4 km. Not km/h. Keep the units straight. They are your friend.
     
  12. Mar 10, 2007 #11
    Okay thanks!
     
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