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Homework Help: Maximum Speed Question

  1. Aug 15, 2003 #1
    I've been trying for about an hour to figure this one out and decided that I should probably ask some people who know what their doing for some input.

    "A driver of a car suddenly sees the lights of a barrier 40 m ahead; it takes the drive 0.75 s to apply the brakes; the car accelerates at a speed of -10 m/s^2. What is the maximum speed possible?"

    I decided to break the problems into two parts; first, the total distance covered during the time prior to initiating of braking and the time after. For the former part I was left with (Vo)(0.75). The latter part I ended with Vo(Vo/10)+0.5(-10)(Vo)^2. I put those together and 40=(Vo)(0.75)+Vo(Vo/10)+0.5(-10)(Vo)^2. Did I create an incorrect equation?
     
  2. jcsd
  3. Aug 15, 2003 #2

    FZ+

    User Avatar

    Welcome to PF!

    I think the second part of the equation is incorrect.

    You are trying to use s = ut + 0.5 a t^2, right?

    You shouldn't as that is 1 equation with two unknowns - you don't know what s is, and you don't know what t is.

    Instead, use v^2 = u^2 + 2*a*s, and seen where you get from there.

    PS: You know SUVAT terminology, right?
    s = displacement
    u = initial velocity
    v = final velocity
    a = acceleration
    t = time
     
  4. Aug 15, 2003 #3
    Thank you for your prompt reply and assistance!

    I changed my equation into v=u^2+2(a)(s) and it worked very well. 0=u^2+2(-10)(40-0.75u)] worked perfectly and I got an answer of 21.76 m/s, or about 78 km/h.

    When the time comes that I'm only 40 meters away from an inescapable barrier I'll know if I'm doomed or not...
     
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