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Velocity, distance, etc.

  1. Aug 28, 2006 #1
    i have the answers.. but just want to make sure im on the right track and didnt mix anything up.

    "a cyclist rode east for 100m with a constant velocity of 25m/s. she then accelerated down a hill and 5 seconds later reached the bottom of the hill with a velocity of 50m/s."

    a) how long did the cyclist ride before reaching the hill?

    100/25 = 4 seconds

    b) what was her average acceleration down the hill?
    (50m/s -25.0 m/s)/5s
    = 5 m/s ^2

    *not sure if the 25 m/s should have been left out of the answer or not*

    c) what was the distance she travelled down the hill?
    5s * 50m/s
    = 250 m

    d) if she continued to ride at constant velocity for another 5s, how far beyond the hill would she travel?
    5m * 25 m/s
    = 125m

    any help will be appreciated!

    ~Amy
     
  2. jcsd
  3. Aug 28, 2006 #2
    a,b - correct
    c,d - wrong
    Gotta go, I have class now.
     
  4. Aug 28, 2006 #3
    (a) and (b) are correct. :wink:

    (c). To find the distance travelled down the hill you should use the following equation:
    distance = average velocity x time

    (d) displacement = velocity x time
    Why did you use 25 m/s for velocity if the cyclist reaches the bottom of the hill with a velocity of 50 m/s? :grumpy:
     
  5. Aug 28, 2006 #4
    thanks guys!

    for c) i now have:
    0.5(50+25)(5s)
    =187.5 m

    or should i also include the 4 seconds that i calculated in a)?

    for d) i used the 25m/s because in the question it was called the "constant velocity". :cry:

    so for d) should i use 0.5(50+25) (the average) or just 50m/s for the velocity? since she's travelling "beyong the hill" (straight probably - not downhill) she's probably moving slower than 50 m/s as she was going down the hill. imo, she'd be going the same velocity as she went she was just biking TO the hill which was 25m/s.

    correct me if im wrong. :tongue:

    ~Amy
     
  6. Aug 28, 2006 #5
    Your answer is correct. You don't include the 4 second calculated in a) because you are calculating the distance she travelled down the hill so the time calculated in a) as nothing to do with it.

    The velocity is constant only before the cyclist went down the hill. When he is travelling down the hill, he has acceleration. Thus, the velocity cannot be constant.

    You use 50 m/s for the velocity, which is both the instantaneous and average velocity, because, after she reached the bottom of the hill, it is constant.

    I can see that you are messing things a little. Be careful with that, please. :smile:
     
  7. Aug 29, 2006 #6
    thanks for the tips! :biggrin:

    another one:

    if a car is accelerating 4 m/s ^2, how far of distance does it travel after 10 seconds?

    the formula im using is (0.5)(4^2)(10)^2 = 800 m.

    is that right? :shy: or should i leave out the 0.5 so it'd equal 1600 m?

    ~Amy
     
  8. Aug 29, 2006 #7
    My pleasure. :smile:

    You may use the following equation:
    [tex]x(t) = v_0t + \frac{1}{2}at^2[/tex]

    It appears that:
    v0= 0 m/s
    a = 4 m/s2

    Why should you leave out 0.5? Why should it be equal to 1600 m? :shy:
     
  9. Aug 29, 2006 #8
    thanks again! i was using the formula you mentioned, but dont understand why the 0.5 is there. :blushing:

    im finally on the right track :smile:

    more to the question:

    "the car starts from rest and accelerates uniformly at 4 m/s east. at the same instant, a truck traveling with a constant velocity of 90 km/hr east overtakes and passes the car".

    c) the car passes the truck at a distance of 312.5 m beyond the starting point. how fast is the car traveling at that instant?


    so the truck is traveling at 25 m/s (according to my calculation)

    would the formula be:
    (25m^2) + 2(4^2)(312.5)
    = 10625 square rooted
    = 103.08 m/s
    = 10.15 m/s ^2?

    ~Amy
     
    Last edited: Aug 29, 2006
  10. Aug 29, 2006 #9
    for d) the question is how long does the car take to pass the truck?

    4.8 seconds?

    and for e) both drivers suddenly see a barrier 100m away and hit their brakes at the same time. assuming both vehicles decelerate uniformly, and they take 3 seconds to stop, will they stop in time?

    :yuck:

    ~Amy

    for e) i calculated the acceleration at 8.3 m/s for the truck, and 34.33 m/s for the car.

    and for both used the formula:
    d = (v)(t) + (0.5)(a)(t^2).

    for the truck i got 112.48 m

    and for the car i got 463.5 m

    i dunno.. any help will be appreciated

    ~Amy
     
    Last edited: Aug 29, 2006
  11. Aug 30, 2006 #10
    Do you know how to find/work out that formula?

    You correctly used the formula [itex]v_f^2 = v_i^2 + 2a \Delta x[/itex] :approve:. But you made some mistakes when calculating. Would you look at that again, please? :smile:
     
  12. Aug 30, 2006 #11
    Unfortunetely, no.
    This formula may be useful:
    [tex]x(t) = (1/2)at^2[/tex]

    You know the values of:
    x(t) and a
    Thus, you can find t.

    How did you calculated those accelerations? :rolleyes:
    Also, be careful with units. Acceleration is expressed in m/s2 (SI units).
     
  13. Aug 30, 2006 #12
    i checked it again but got the same answer :yuck:

    25/3 = 8.333 (truck)
    103/3 = 34.33 (car)

    im going to look into it. but im not very mathimatical.


    what variable is x(t)?
    just to be clear:
    the "a" = acceleration
    "t" = time in seconds?

    thanks for the help! i really appreciate it :smile:

    ~Amy
     
  14. Aug 30, 2006 #13
    update!

    for c) i figured it out the old fashion way (creating a graph) to find the time for 312.5 (based on acceleration of 4 m/s^2) and ended up with 50 m/s. i hope that's right!

    ~Amy
     
  15. Aug 30, 2006 #14
    im still having trouble with c) :confused:
    my current calculation is (4.0^2)(12.5) = 200 m/s.. but that doesnt sound right or is it?

    and for d) i got 12.5 seconds

    ~Amy
     
  16. Aug 30, 2006 #15
    The .5 is there because of the relationship between acceleration, velocity, and position.

    velocity is the derivitive of position.

    x = x0 + v0t + 1/2at^2
    x is position
    v = v0 + at
    v is velocity

    so the 1/2 is there because t is squared, and it's needed to offset the 2 when you diferentiate.

    might not be too clear, but I hope that helped
     
  17. Aug 30, 2006 #16
    thanks bigcheese!

    does anyone have the foggiest idea how to solve this one?
    without knowing the acceleration rates, is it even possible?

    and for c) my latest calculation is 100m/s.. it really depends what formula i use :uhh:

    ~Amy
     
  18. Aug 30, 2006 #17
    in the equation

    x = x0 + v0t + .5at^2

    remember that a is going to be a negative value, because they're decelerating. See if that helps at all.
     
  19. Aug 30, 2006 #18
    ah, also, this just occured to me. if they go from v to 0 uniformly, and it takes them 3 seconds to do that, all you need is v to determine their accel.

    a = change in velocity, so if they're going 90, and in 3 secs they go 0, their accel would be -30m/s^2
     
  20. Aug 31, 2006 #19
    update:

    for a) i got 200 m/s (my original answer of 800 m was wrong)
    c) 50 m/s
    d) 12.5s
    e) 37.65 m for the truck, and 75 m for the car. for deceleration i got -8.3m/s^2 for the truck, and -16.67 m/s^2 for the car.

    :smile:

    thanks thebigcheeze and PPonte!

    ~Amy
     
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