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Acceleration and velocity question

  1. Apr 24, 2005 #1
    hey, i have a lab due for a grade 10 physics class tomorrow. Im not too smart with this stuff, and I'm stuck with this question. can someone please help me out;

    Question: how long would it take a cart on a similar ramp to reach a velocity of 90km/h?

    i did a lab where we measured the speeds of a cart going down a ramp with a ticker tape timer thing.

    i calculated that the average velocity was 166cm/s [down]

    ........... theres one more question too;

    calculate the velocity the cart would have if it was allowed to run down a similar ramp for a period of 5.0s


    im sure this is real basic stuff, im just stuck

    my data was

    time 0.0 velocity (cm/s) 0
    time 0.05 velocity 16
    time 0.15 velocity 31
    time 0.25 velocity 47
    time 0.35 velocity 63
    time 0.45 velocity 72
    time 0.55 velocity 99
    time o.65 velocity 107
    time 0.75 velocity 123
    time 0.85 velocity 183
    time 0.95 velocity 154

    its basically uniform when plotted on a velocity vs time graph
     
  2. jcsd
  3. Apr 24, 2005 #2

    OlderDan

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    Your graph has all the information you need. Draw the best straight line you can through your data and extrapolate (extend) your graph to longer times and higher velocities if needed. For the first question, you need to convert from km/hr to cm/sec. At what time does the graph cross 90km/hr? (Average velocity is not the most direct approach to answer this question. The number you have for average velocity is not what the data says. Are you sure that is velocity?)

    For the seond question, what velocity is on your graph at t = 5 sec?

    Extrapolation of the graph can be done in your mind if not done on paper. What is the slope of the graph? How can you use that to decide where points would be at larger times and larger velocities?
     
  4. Apr 24, 2005 #3
    Try and get an equation for the graph and everything should be easy from there.
     
  5. Apr 24, 2005 #4
    im still stuck, i understand what you're saying. but i dont know how to do it. i dont know proper equations, thats why im on these forums
     
  6. Apr 24, 2005 #5

    OlderDan

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    Your data graph is "uniform" you said. I take that to mean a straight line. The general equation of a line is

    y = mx + b

    In your case y is velocity, and x is time. b is the value of y when x is zero, which is zero for you. The equation that represents your line is

    v = mt

    and you calculated the slope to be 166m/s^2, except that you did not get the units right. We call the change in velocity divided by the change in time acceleration, and that is the slope of your graph. Your equation is

    v = (166cm/s^2)t

    Using that equation, you can figure out what time gives a velocity of 90km/hr, but you need that velocity expressed in cm/sec. The best way to convert units is to multiply what you have by fractions that are equal to one that replace unwanted units with wanted units.

    [tex]90\frac{km}{hr}=90\frac{km}{hr}\left[\frac{1000m}{km}\right]\left[\frac{100cm}{m}\right]\left[\frac{1hr}{60min}\right]\left[\frac{1min}{60sec}\right]= how-many\left[\frac{cm}{sec}\right][/tex]

    Plug the result into the equation relating v and t and solve for t.

    For the last question, plug in for t to find v.
     
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