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Homework Help: A car travelling at 50km/h stops in 70.0m HELP

  1. Mar 8, 2014 #1
    1. The problem statement, all variables and given/known data

    A car travelling at 50.0 km/h stops in 70.0 m. What is the stopping distance if the car's speed is 90 km/h?

    2. Relevant equations

    This is a question in the unit about momentum & energy, so most of the questions have been applying formulae about momentum, work, gravitational potential energy, kinetic energy, etc. I'm confused as to how this question relates to this section because I can't seem to find an applicable formula.

    3. The attempt at a solution

    I attempted to solve this using ratios and got an answer of 126 m, but the real answer is somewhere around 2.3E2 m. I'm super confused, but expecting that it'll be a super obvious solution...?
  2. jcsd
  3. Mar 8, 2014 #2
    OH! Maybe I figured it out. Since doubling velocity QUADRUPLES the stopping distance (because 2^2 = 4), and the velocity was increased by 1.8, I can square 1.8 (1.8^2 = 3.24) and multiply 3.24 by 70.0 to get 226.8! Does anyone know how I could show this in a more mathematical way?
  4. Mar 8, 2014 #3

    Simon Bridge

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    You can also see it in terms of the v-t graph.
    Model the car as stopping with a constant acceleration ... this is a line on your graph with a negative slope - so it makes a triangle with the v and t axes. The area inside this triangle is the distance to stop.

    Sketch the graph for both situations, then you have two similar traingles - don't let the time to stop T being unknown phase you - what's important is that the acceleration is the same each time.

    If you prefer to memorize equations - the equations you want are the kinematic or "suvat" equations.
    You want the one that relates displacement to velocity and acceleration.
  5. Mar 9, 2014 #4


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    I was going to suggest writing two energy equations using..

    Work = Force * distance = 0.5mV2


    F * D1 = 0.5mV12

    F * D2 = 0.5mV22

    Substitute for F.
    The mass will cancel.
  6. Mar 9, 2014 #5


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    I like your approach here.
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