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Stopping Distance Grade 11 Physics Help Needed -

  1. Oct 25, 2009 #1
    Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    Yatin was driving at 90.0km/h when he saw the light turn yellow at an intersection located 75.0m ahead. Yatin used 0.400s to make up his mind, and then braked at 5.00m/s2 until he stopped. What was his stopping distance? Did Yatin make the right decision? If not, what could he have done differently?



    (ATTEMPT BELOW COPIED/PASTED FROM MS WORD EQUATION EDITOR [MY WORK])
    v1=90.0km/h or 25.0m/s
    a=-5.00/s/s
    v2=0.00m/s
    Step 1:
    t=v1/a
    t=25.0/5.00
    t=5.00s
    Step 2:
    d after making decision=d-(v1*t)
    d after making decision=75-(25*.04)
    d after making decision=75-1
    d after making decision=74.0m

    Step 3:
    stopping distance=dtotal-(a*t)
    stopping distance=74m-(5*5)
    stopping distance=74-25
    stopping distance=49m
    Therefore Yatin`s stopping distance is 49m
     
  2. jcsd
  3. Oct 25, 2009 #2

    Delphi51

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    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    Hello Newb, welcome to PF!
    In step 2, you used a time of .04 s when the question said 0.4 s.
    In step 3, you used a*t as a distance, but in fact it is a velocity.
    It seems to me the solution would be clearer if you just calculated the distance for each of the two motions, then add them up and see if they exceed the 75 m.
     
  4. Oct 25, 2009 #3
    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    Thanks delphi...i think i fixed it:
    Yatin was driving at 90.0km/h when he saw the light turn yellow at an intersection located 75.0m ahead. Yatin used 0.400s to make up his mind, and then braked at 5.00m/s2 until he stopped. What was his stopping distance? Did Yatin make the right decision? If not, what could he have done differently?
    v1=90.0km/h or 25.0m/s
    a=-5.00/s/s
    v2=0.00m/s
    Step 1:
    t=v1/a
    t=25.0/5.00
    t=5.00s
    Step 2:
    d after making decision=d-(v1*t)
    d after making decision=75-(25*.400)
    d after making decision=75-10
    d after making decision=65.0m

    Step 3:
    stopping distance=dtotal-[(a*t)*t]
    stopping distance=65m-(5*5)*5
    stopping distance=65-25*5
    stopping distance=65-125
    stopping distance=-60m
    Therefore Yatin`s stopping distance is 60m after the light.


    Is this right? I didn't understand what you meant by calculating the distance of both motions?
     
  5. Oct 25, 2009 #4
    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    Would really appreciate help...The thing is due in a couple hours...i have 2 more questions to do :S...tensed!!
     
  6. Oct 25, 2009 #5

    Delphi51

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    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    Step 3 - you used a*t*t for the distance, but this still is not the correct formula for distance. The negative distance means the car has reversed direction - a clear indication that something is wrong in the calculation.
     
  7. Oct 25, 2009 #6
    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    These were the only formula's given to us....would you be able to tell me which one to use in this case?
    http://img196.imageshack.us/img196/7408/captureie.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. Oct 25, 2009 #7

    Delphi51

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    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    Any of the ones with a D in them should work, if you can figure out the stopping time first.
     
  9. Oct 25, 2009 #8
    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    wait so do you mean the stopping time i calculated was wrong?
    Step 1:
    t=v1/a
    t=25.0/5.00
    t=5.00s
     
  10. Oct 25, 2009 #9

    Delphi51

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    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    Oh, I forgot you had that. It is correct!
     
  11. Oct 25, 2009 #10
    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    ok i think i got it now....is this correct?
    v1=90.0km/h or 25.0m/s
    a=-5.00/s/s
    v2=0.00m/s
    Step 1:
    t=v1/a
    t=25.0/5.00
    t=5.00s
    Step 2:
    d after making decision=d-(v1*t)
    d after making decision=75-(25*.400)
    d after making decision=75-10
    d after making decision=65.0m

    Step 3:
    stopping distance=d- [(v1+v2)/2] t
    stopping distance=65-[(25+0)/2]*t
    stopping distance=65-(12.5*5)
    stopping distance=65-62.5
    stopping distance=2.5m
    Therefore Yatin`s stopping distance is 2.5m before the light.
     
  12. Oct 25, 2009 #11

    Delphi51

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    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    Got it!
    Would perhaps be nicer if you wrote
    d = vt + [(v1+v2)/2] t
    = 25*0.4 + (25 + 0)/2*5
    = 72.5 m
     
  13. Oct 25, 2009 #12
    Re: Stopping Distance Grade 11 Physics Help Needed -- URGENT!!

    Got it Thanks
     
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