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Motion questions

  1. Jul 13, 2008 #1
    1. The problem statement, all variables and given/known data
    You are diving to the grocery store at 20 m/s. You are 110 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.70 s and that your car brakes with constant acceleration.

    A) How far are you from the intersection when you begin to apply the brakes?
    B) What acceleration will bring you to rest right at the intersection?
    C)How long does it take you to stop?


    2. Relevant equations
    Xf = Xi + ((Vx)i)(Delta T) + .5(Ax)(Delta T)^2


    3. The attempt at a solution

    This may be a pretty basic problem for you folks here. But I honestly don't have the slightest clue where to even start. I'd imagine id have to find the acceleration of the vehicle in order to find part A. But I don't see how to find it. Ax= Delta Vx/ Delta T correct? So with the information given would it be Ax= 20m/s / 0.70 s ?

    I have the answers to the problem from my text book. I Just don't understand the steps in solving for the answer. If anybody can give me a bump in the right direction I would appreciate it. Thanks.

    Oh yeah, Answers are

    A) 10 m
    B) -2 m/s^2
    C) 11 sec
     
  2. jcsd
  3. Jul 13, 2008 #2

    tiny-tim

    User Avatar
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    Homework Helper

    Welcome to PF!

    Hi osker246! Welcome to PF! :smile:

    (I do nudges rather than bumps … :smile:)

    Part A is just distance = speed x time … you don't need acceleration at all! :smile:

    In fact, look at your equation:
    (m/s) / s would be m/s2, wouldn't it? :wink:

    Always check the dimensions of a formula, if you're not sure. :smile:
     
  4. Jul 14, 2008 #3
    Re: Welcome to PF!

    Thank you for the welcome!

    I don't know if D=RT is a proper application to this problem though. The only time given in this problem is .70 sec which refers reaction time not travel time. If I'm incorrect please do inform me. I'm just a bit lost where to go with this problem.
     
  5. Jul 14, 2008 #4
    ok im just messing around with numbers right now.

    So 110 m / 20m/s = 5.5 sec

    So if the vehicle continues at a rate of 20m/s it will take 5.5 sec to travel the 110 m correct?

    So finding the acceleration should be as easy as 20 m/s / 5.5 s = 3.63 m/s^2. Right?

    So would I go about using this equation?

    Xf = Xi + ((Vx)i)(Delta T) + .5(Ax)(Delta T)^2

    xf=110 m
    (Vx)i=20 m/s
    Delta T= .70 sec?
    Ax= 3.63

    Then solve for its initial position?
     
  6. Jul 14, 2008 #5
    yeah....I have no idea what the hell I'm doing....
     
  7. Jul 14, 2008 #6
    Re: Welcome to PF!

    Yea, you can use d = rt for this because that means that for .7 secs you're still traveling at a constant rate of 20 m/s.

    For B, use the distance you found in A; remember, you also know the initial velocity, the final velocity (what equation can you use?)

    For C, you know acceleration, initial velocity and final velocity (what equation will work here?)
     
  8. Jul 14, 2008 #7
    Re: Welcome to PF!

    So...

    D=(.70)(20)=14 m

    for that .70 of a second the vehicle traveled 14 m. So 110m - 14m = 96m

    So does that mean I'm 96m away from the intersection before starting to slow down?
     
  9. Jul 14, 2008 #8
    Let the distance from the intersection, when the driver sees the red light be [tex]L[/tex], let his rection time be [tex]t_{0}[/tex]. Let the initial speed of the car be [tex]v_{0}[/tex]. If the driver decides to put on the brakes the moment he sees the red light, the action still takes time [tex]t_{0}[/tex], and in that time the car has traveled a distance [tex]v_{0}t_{0}[/tex]. Hence the car travels with deceleration for the distance [tex]L-v_{0}t_{0}[/tex]. This is the answer to (a)
    If [tex]a[/tex] be the magnitude of the deceleration, then
    [tex]0=v^{2}_{0}-2a(L-v_{0}t_{0})[/tex]
    which gives [tex]a[/tex], the answer to (b)
    The time can be found from
    [tex]L-v_{0}t_{0}=v_{0}T-\frac{1}{2}aT^{2}[/tex]
    which gives the time of deceleration. The time of stopping will be [tex]T+t_{0}[/tex], which gives the answer to (c)
     
  10. Jul 14, 2008 #9
    Re: Welcome to PF!

    Correct
     
  11. Jul 14, 2008 #10

    tiny-tim

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    reaction time

    Hi osker246! :smile:

    (I've been away all day, but I see cryptoguy :smile: has helped you to the right result for A).)
    Time is time. The .70 sec is both reaction time and travel time.

    Reaction time (you don't drive do you?) means the time it takes someone (usually between .60 and 1.5 sec, I think) to react after they see something … as when you see someone lying in the road, and you "immediately" slam on the brakes, but it takes .70 sec for the message to get from you eyes to your brain, and then back to your foot!

    In other words: it takes you .70 sec just to "react", before you actually do anything! :smile:

    (If you're not sure what a word means, just ask! :wink:)

    How are you doing on B) and C)? :smile:
     
  12. Jul 14, 2008 #11
    I got the problem solved. Turned out the teacher handed out the wrong answer for part A. This is why I had so much trouble trying to solve this problem. My answers were not matching the one he originally handed out. I just want to say thank you for the help everybody! Its greatly appreciated.
     
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