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Constant acceleration of sports car

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data


    You are driving your new sports car at a velocity of 90km/hr , when you suddenly see a dog step into the road 50m ahead . You hit the breaks hard to get maximum deceleration of 7.5m/s² .



    2. Relevant equations

    i)How far will you go before stopping?

    ii)Can you avoid hitting the dog?

    can anyone help me ? please ?
     
  2. jcsd
  3. Apr 25, 2012 #2
    What equations do you know that involve kinematics?
     
  4. Apr 25, 2012 #3
    the topic was constant acceleration .. then i hardly to understand .. it means i cant understand my teacher .. not only me .. >.<
     
  5. Apr 25, 2012 #4
    Well the kinematic equations can be found on this webpage. Then read the paragraph that follows the equations, and try to understand them well because you'll need to use them a lot. These equations ONLY apply when acceleration is constant.

    For this question you are told the:
    Initial velocity, vi = 90 km/h
    Final velocity, vf = 0 km/h (because you've stopped)
    Acceleration, a = -7.5 m/s^2

    So look for an equation that has all three of these values in order to get something useful :). Be careful with the units, and make sure you are consistent with length and time (i.e. make sure you always use meters and seconds, or kilometers and hours).
     
  6. Apr 25, 2012 #5
    my answer was -150m .. and the true answer was 42m only ..

    I think my brain will be explode ..

    can you show how to solve EmittingLight ?

    please ?
     
  7. Apr 25, 2012 #6
    It's better if you show your working so that we can show where you went wrong.

    I suspect it might be unit conversions, which you can check with tools like Wolfram Alpha. The initial velocity is 90 km/h whereas the acceleration uses meters and seconds so you have to convert one of them to the other's units. I like using SI units, so I'd convert to meters and seconds. In Wolfram Alpha you would type something like this. With all the unit conversions you should get:

    Initial velocity, vi = 25 m/s
    Final velocity, vf = 0 m/s (because you've stopped)
    Acceleration, a = -7.5 m/s^2

    If you look at the equations there are only two that have all of these variables:
    [tex]v_{f}^2=v_{i}^2+2*a*d[/tex][tex]v_{f}=v_{i}+a*t[/tex]You know how to solve for t and d in the equations right? t stands for the time elapsed and d stands for the distance traveled. Do you know which equation you want then?
     
  8. Apr 25, 2012 #7
    to solve for t:

    i used the second eq.

    for d:

    i used the first eq .

    my t is -17.5s

    then to my d is -610m

    then I divide the both to have velocity ..

    what's next ? :?
     
  9. Apr 25, 2012 #8
    Three of the values are known, I've told you what they are in all my previous posts: [itex]v_f, v_i[/itex] and [itex]a[/itex].

    I'll put these into the equations for you:[tex]0^2=25^2+2*-7.5*d[/tex][tex]0=25+-7.5*t[/tex]Do you know how to solve for t and d now? Remember, they are all in meters and seconds now, since that's what I've converted them into.

    Would you be able to show your working? Like show how you solved for t and d, so I can tell you what you did incorrectly.

    EDIT: Nevermind I know what you're doing. You're not doing the algebra correctly. Do you know that if:
    [itex]a=b*x[/itex]
    That:
    [itex]\frac{a}{b}=x[/itex]?

    So if:
    [itex]-625=2*-7.5*d[/itex]
    Then:
    [itex]\frac{-625}{2*-7.5}=d[/itex]
     
    Last edited: Apr 25, 2012
  10. Apr 25, 2012 #9
    0^2=25^2+2∗−7.5∗d
    0=625-15d
    (transfer:)
    15d=625
    divide both sides by 15 to remain the d
    then, d=41.67m

    for the second:
    0=25+-7.5*t
    0=25-7.5t
    (transfer:)
    7.5t=25
    divide both sides by 7.5 to remain t:
    t=10/3s

    is it wrong ?
     
  11. Apr 25, 2012 #10
    Well, didn't you say that the true answer was 42m before? It looks like it was rounded up from 41.67m to me.

    Do you know why you got t=-17.5 and d=-610 before? And not now? Based on what I've deduced, you actually already put in the correct values before.

    EDIT:
    I just thought I should add that since you only wanted d, you didn't need to use the second equation. That's why I asked before if you 'knew which equation to use'; you wanted to find d, so you needed to use the first equation. But look on the bright side of this extra confusion, now you know how long it took the car to stop! And yes, you have the correct answers now as far as I know.
     
    Last edited: Apr 25, 2012
  12. Apr 25, 2012 #11
    I think i'm going to crazy .. but so very thank you .. now i know how the constant acceleration ..
     
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