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What is the final velocity?

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data

    a car is at a velocity of 20 km/h if the car traveled 120 km in 3 hours at constant accelration. what is the final velocity?

    2. Relevant equations


    V^2=Vo^2+2ad

    3. The attempt at a solution

    I am completely confused. The book does not even list the particular formula above. And not much practice examples are given . I have an issue with understanding these variables. I dont see the correlation between the variables and the particular numbers. and its making a potentially easy problem non understandable.

    please help
     
  2. jcsd
  3. Sep 28, 2011 #2

    cepheid

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    Re: Kinematics-velocity&acceleration

    Welcome to PF albodibran!

    In the equation you typed above, v is the current velocity (when the object is at distance d from its starting point), v0 is the initial velocity (when the object began accelerating). The variable 'a' is the acceleration (which is constant in this case) and d is the total displacement.

    However, you don't need this equation. In fact, you don't even need the distance value you were given. Answer this question: what is the definition of acceleration?

    Bearing in mind the definition above, if you have an acceleration value, and a time interval, what can those two things tell you about the velocity over that time interval?
     
  4. Sep 28, 2011 #3
    Re: Kinematics-velocity&acceleration

    First, welcome to PF!

    Let's see if we can help get you straightened out.

    If we have constant acceleration,

    [itex] V= V_0 + at (1)[/itex]

    [itex]V[/itex] is the final velocity, [itex]V_0[/itex] is the initial velocity, [itex]a[/itex] is the acceleration, and [itex]t[/itex] is the time.

    Also, we have the displacement,
    [itex] d = \frac{1}{2}(V + V_0) t (2)[/itex]

    [itex]d[/itex] here, as said, is the distance traveled, also known as displacement.

    If we rearrange (1) to solve for t, we get
    [itex] t = \frac{V-V_0}{a} (3)[/itex]

    Now from here, just substitute (3) into (2) for t. You then have:
    [itex]d = \frac{1}{2}(V + V_0) (\frac{V-V_0}{a})[/itex]

    When you work out your multiplication (I hope you're verifying what I'm doing here, it'll help) and rearrange things,

    [itex]V^2 = V_0^2 + 2ad[/itex]

    Does that help you to see where they are coming from? If not, let us know and we'll see what we can do for you.
     
  5. Sep 28, 2011 #4
    Re: Kinematics-velocity&acceleration

    I figured that out thank you I will post a issue I have with part of another problem
     
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