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Deriving Basic Physic Problem Using Calculus

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data

    At a distance of 45 meters from a traffic light, a car traveling 15 m/s is brought to a stop at a constant deceleration

    a) What is the value of Deceleration (-5m/s)
    b) How far has the car moved when its speed has been reduced to 3 m/s
    c) How many seconds would the car take to come to a full stop?

    2. Relevant equations

    I really want to solve this Physics based, but we need to use calculus to derive the equations. so there's no way of showing what equations i have used already

    3. The attempt at a solution

    I don't have an attempt at b and c because i don't know how to start them.
    Help would be appreciated, it's the last two questions i have to do on an assignment for tomorrow. Please help
     
  2. jcsd
  3. Sep 22, 2011 #2

    dynamicsolo

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    The acceleration function for the car is a(t) = -5 , a constant function. What are the velocity and position functions for this? (What is the relationship between a(t) and v(t)? between v(t) and x(t) ?
     
  4. Sep 22, 2011 #3
    the relationship between a(t) and v(t), and v(t) and x(t) is that if you integrate a(t) you get v(t) and integrating v(t) gets x(t)
     
  5. Sep 22, 2011 #4

    dynamicsolo

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    Good! So [itex]v(t) = \int a(t) dt = \int (-5) dt [/itex] gives you...?
     
  6. Sep 22, 2011 #5
    Well, from what i can tell, it give you the speed at any given time.
    But if that's the case i don't have the time to solve B
     
  7. Sep 22, 2011 #6

    dynamicsolo

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    You get an "arbitrary constant" from the integration. What you know from the statement of the problem is that when the car begins slowing down, which is at t = 0, the speed is v(0) = 15 m/sec. What does this make the constant from your velocity integral?
     
  8. Sep 22, 2011 #7
    Acceleration???
    Sorry, this really isn't clicking.
     
  9. Sep 22, 2011 #8

    Ray Vickson

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    You might find it easier to consider the problem in reverse: starting from rest, a car undergoing constant acceleration reaches a speed of 15 m/s at a distance of 45 m.

    RGV
     
  10. Sep 22, 2011 #9

    dynamicsolo

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    Are you asking about the "-5" ? Acceleration is any change in velocity, so slowing down is a negative acceleration (what people also called a "deceleration"). The calculations of motion can be done with positive or negative accelerations; the ability to work out the velocity and position functions does not depend on the sign of a(t).
     
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