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Suvat equations - car braking hard to avoid a collision

  1. Jan 24, 2014 #1
    Suvat equations -- car braking hard to avoid a collision

    1. The problem statement, all variables and given/known data

    A car (A) is speeding along a rural road, of width 4m with a speed of 25m/s when the driver sees another vehicle (B) of length 3.5m just starting to cross the road at a point 40m ahead. The drivers reaction time is 0.80s, and the maximum deceleration with the brakes fully applied is 5.5m/s/s, If the car can be assumed to decelerate uniformly without swerving etc., calculate:

    a) The distance travelled by the car during the drivers reaction time
    b) The car's (A) velocity when it reaches the position of the other vehicle (B)
    c) Total time which has elapsed from first sighting until car A reaches vehicle B
    d) The minimum constant velocity of vehicle B so that the car A does not collide with it

    2. Relevant equations

    SUVAT Equations

    3. The attempt at a solution

    a)
    s=
    u=25m/s
    v=
    a=-5.5m/s/s
    t=0.80

    s=ut+0.5(at2)
    s=18.24m

    b)
    s=40m
    u=25m/s
    v=
    a=-5.5m/s/s
    t=

    v2=u2+2as
    v2=252+2(-5.5)(40)
    v=13.60m/s

    c)
    s=40m
    u=25m/s
    v=13.60m/s
    a=-5.5m/s/s
    t=

    v=u+at
    13.60=25-5.5t
    t=(13.60-25)/-5.5=2.07s

    d) Rather unsure on this one:
    s= 3.5m
    u=
    v=
    a=0m/s/s
    t=2.07s

    I never calculated anything for d) because I am really unsure.

    Any help is appreciated!
     
  2. jcsd
  3. Jan 24, 2014 #2
    Reaction time is the time taken by the driver to start applying brakes on seeing the obstacle(car B in this case) .During the reaction time the car A is travelling at 25m/s with no deceleration.

    The item in red should be 20m
     
    Last edited: Jan 24, 2014
  4. Jan 24, 2014 #3
    Ok therefore:

    a)
    s=
    u=25m/s
    v=
    a=0m/s/s
    t=0.80

    s=ut+0.5(at2)
    s=25(0.80)+0
    s=20m

    Subsequently for b)

    b)
    s=40-20=20m
    u=25m/s
    v=
    a=-5.5m/s/s
    t=

    v2=u2+2as
    v2=252+2(-5.5)(20)
    v=20.1m/s

    And for c)
    c)
    s=20m
    u=25m/s
    v=20.1m/s
    a=-5.5m/s/s
    t=

    v=u+at
    20.1-25/-5.5=t=0.89s
    So total t=0.89+0.80=1.69s

    Is this right?
     
  5. Jan 24, 2014 #4
    Yes...that's right
     
  6. Jan 24, 2014 #5
    Now comes part d)

    My thinking is:
    s=3.5m
    u=
    v=
    a=0m/s/s
    t=1.69s

    Then work out u?
     
  7. Jan 24, 2014 #6
    No...

    How much distance car B needs to travel to completely cross the road ?
     
  8. Jan 24, 2014 #7
    4 metres
     
  9. Jan 24, 2014 #8

    adjacent

    User Avatar
    Gold Member

    Nope.The car has to COMPLETELY cross the road.The width of the road is 4m.Length of the car is 3.5m
    Think about it.
     
  10. Jan 24, 2014 #9
    7.5 metres?
     
  11. Jan 24, 2014 #10
    Car is not a particle .It has a length .The distance to be covered is the width of the road + length of the car .
     
  12. Jan 24, 2014 #11
    Yes...
     
  13. Jan 24, 2014 #12
    So am I right to say:
    d)
    s=7.5m
    u=
    v=
    a=0m/s/s
    t=1.69s

    s=ut+0.5(at^2)
    50=1.69u
    u=50/1.69=29.6m/s
     
  14. Jan 24, 2014 #13
    :rolleyes: s should be 7.5 not 50 .
     
  15. Jan 24, 2014 #14
    Sorry I was looking at my maths homework just before hand, and 50 was the answer to a question, wasn't paying attention :shy:
     
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