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Motion in a Straight Line

  1. Dec 15, 2006 #1
    1. The problem statement, all variables and given/known data

    I'm taking a grade 11 physics correspondence course and would like to know if I am on the right track.

    At the instant a tracffic light turns green, a car starts from rest and acccelerates uniformly at a rate of 4.0m/s. At the same instant, a truck travelling with a constant velocity of 90.0 km/hour overtakes and passes the car.

    a. how far beyond the starting point is the car after 10.0 s?
    b. how far beyond the starting point is the truck after 10 s?
    c. The car passes the truck at a distance of 312.5 m beyond the starting point. How fast is the car travelling at this instant?

    2. Relevant equations

    Equations i'm using are: distance =(vi)(t) + (0.5)(a)(t^2)
    Speed =distance/time

    3. The attempt at a solution

    Here is what I have so far.

    a. d=(vi)(t) + (0.5)(a)(t^2)
    =(0m/s)(10s) + (0.5)(4.0m/s^2)(10s^2)
    =0.5(400)
    =200 m for the car after 10 s

    b. distance = (constant velocity) (time)
    =(25m/s)(10s)
    = 250 m for the truck after 10 s

    c. I have worked out a calculation but my answer doesn't make sense
    200m/10s=312.5m/?s
    =15.6 s for the car to go 312.5 m
    from this i calculated the speed
    speed=distance/time
    =312.5m/15.6 s
    =20 m/s then i translated into km and got 72 km/hour. However I know this is wrong because logically if the car is passing the truck it would be a number greater than 90.0 km/h. What concept am I missing here???
    Thanks for your help.
    Pharm 89
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 15, 2006 #2

    berkeman

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    Staff: Mentor

    a and b are fine. On c, you are deriving the time incorrectly. Use your relevant equation that you listed:

    "distance =(vi)(t) + (0.5)(a)(t^2)"

    to figure out the time (remember to take the square root when you get to it), and then plug that into the equation for the velocity of the car as a function of time.

    Nice use of the Homework Posting Template, BTW. Showing the relevant equations like that and listing all the steps in your work makes it a lot easier to help you. Welcome to the PF.
     
  4. Dec 16, 2006 #3
    Thank you very much for your help berkeman. I used the distance formula like you said and rearranged it to find the time . Did I rearrange this formula the correct way.
    d= (vi)(t) + 0.5 (a)(t^2)
    t3= (vi) + (0.5) (a) (d)
    =0 + 0.5 (4) (312.5)
    =625
    t = 8.54 s

    then I plugged this answer into the speed formula =distance/time
    = 312.5 m/8.54 s = 36 m/s then convert into km = 129 km/hour

    Wouldn't the answer to the question:how long does the car take to pass the truck? be 8.54 s which i solved in the previous equation.
    Thanks for the assistance.
    Pharm 89
     
  5. Dec 22, 2006 #4
    Im having a little bit of trouble understanding this question. Would anyone be able to tell me if I am thinking this through properly.
    other important pieces of info: a car starts from rest and accelerates uniformly at a rate of 4.0 m/s and the truck travels at a constant velocity of 90.0 km /h
    Both drivers suddenly see a barrier 100.0 m away and hit their brakes at exactly the same time. Assuming that both vehicles decelerate uniformly and they take 3.0 s to stop, will they stop in time??

    I calculated that at a distance of 312.5 m the car is travelling 129 km/hour therfore would you divide that by 3 sec. The car would not stop in time. The truck is going 90 km/h which would equal -30 m/s and therefore would stop in time.
    Thanks for the feedback
    Pharm 89
     
  6. Dec 23, 2006 #5

    PhanthomJay

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    Gold Member

    First, you're having a bit of trouble with your algebra. I like to plug and chug:
    Since
    d= (vi)(t) + 0.5 (a)(t^2), then
    312.5 = 0 + 0.5(4)(t^2)
    312.5 = 2t^2
    156.25 = t^2
    t^2 = 156.25
    t = ????
    Next, having found t, you need the formula that relates instantaneous velocity with time. You chose the wrong one, because distance/time is average velocity, not instantaneous velocity. What is the formula that relates instantaneous velocity with acceleration and time?
     
  7. Dec 23, 2006 #6

    PhanthomJay

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    I'm assuming you mean that at 312.5 meters, with the car and truck now side by side, they see the barrier 100m away and brake, and both stop 3 seconds later. Once you have correctly determined their respective speeds when each hits the brakes, use your relevant equation (as corrected above) average speed =distance/time, but first note that average speed is (v_i + v_f)/2
     
  8. Dec 23, 2006 #7
    Thanks Phanthom Jay for your help in pointing me in the right direction. It makes much more sense to me now. Based on your steers I figured out that the car would be travelling at 50 m/s (180 km/hour) and it would take 12.5 s for the car to pass the truck. The question about both drivers seeing the barrier 100 m away and they both applied brakes at the same time I figured out that the car would be deccelerating at -16.67 m/s and the truck would be deccelerating at -8.33 m/s. Then I plugged this information into the distance formula and got that it would take 75 m for the car to stop and 37.5 m for the truck to stop, so yes they would both stop in time.
    I hope I came up with the right answers.
    Thanks again for the help.
    Pharm 89
     
  9. Dec 24, 2006 #8

    PhanthomJay

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    Yes, you got it all right, nice job. You could have just used the formula
    d = (v_avg)(t), where V_avg for the car is 25m/s ( (50 + 0)/2 = 25 ), and v_avg for the truck is 12.5m/s, and achieved the same result. This avoids having to calculate the deceleration. But there are many ways to solve a problem, and your way works just fine.
     
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