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Homework Help: T_T lots of questions

  1. Jan 27, 2008 #1
    Kinematic Equations... Constant Speed

    1. The problem statement, all variables and given/known data
    Question 1) A car pulls away from an intersection when the light turns green. After uniformly accelerating for the next 4.0s, the car has traveled a distance of 50m. The car then proceeds at constant speed.
    1. What was the car's acceleration
    2. How fast was the car traveleing when it finished accelerating
    3. How long will it take for the car to travel another 50 m at this constant speed?
    2. Relevant equations
    The 5 kinematic equations

    3. The attempt at a solution
    For #1... I used a kinematic equation and I found this info
    Initial Velocity = 0
    Time = 45s
    X = 50
    a = ?

    X = distance
    a = acceleration

    So then I used x=intitial velocity + acceleration*time
    and got 1.11m/s.... Am I right?

    For #2... I found
    Initial Velocity = 0
    Time = 4 seconds
    X = 50 minutes
    a = 1.11m/s
    V= ?

    Would objects position at time = 0 be relevant here?

    I got a answer of 14.72 m/s... Is that right?

    For #3
    I said X= 100 (50 m + 50 m)
    said initial velocity = 0
    a = 1.11m/s
    t = ?
    I got t = 90s... Kind of makes sense, 14.72m/s + 90s = 104 something M...

    ? - the unknown. Please help :D

    Last edited: Jan 27, 2008
  2. jcsd
  3. Jan 27, 2008 #2
    I think you have a problem right away.

    Why are you using 45s as the time? It appears to be 4.0s in your problem. Try recalculating your acceleration.

    I'm not sure what this is, and I think you're mixing your formulas up.
    For this #2, you're close, except you'll have to recalculate using the new acceleration you should have found above. The easiest formula to use to find the velocity at that point is:

    [tex]v_2 = v_1 + at[/tex]
    For #3 below, I think you've really gone astray. It's only asking for the time for the next 50m, and an important piece of information is:
    Remember, a constant speed means NO acceleration. So you can just use a constant velocity equation, such as:
    [tex] v = \frac{dist}{time}[/tex]

    So just toss this stuff below out and try again.
    Good luck!
  4. Jan 29, 2008 #3
    Still cant find the answer to the cars acceleration
  5. Jan 30, 2008 #4
    Well, you need to use the correct equation, which I mentioned above. You have

    X = initial velocity + acceleration * time

    but that's not the correct equation. The correct equation is:

    [tex] \Delta x = ut + \frac{1}{2} a t^2[/tex]

    Where u = initial velocity. You can solve for aceleration using
    [tex] \Delta x = 50m[/tex]
    [tex] u = 0 [/tex]
    [tex] t = 4.0s [/tex]

    So now we have:

    [tex]50 = (0)(4) + \frac{1}{2} a 4^2[/tex]

    You can do the rest.
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