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Stuck on these questions

  1. Jun 17, 2004 #1
    I've been trying to figure them out but cant get through about 6 questions, these are the other two..

    An object is observed to fall from a bridge, striking the water below 2.50 s later. (a) With what velocity did it strike the water? (b) What was its average velocity during the fall? (c) How high is the bridge?

    A ferryboat requires 45.0 min to travel 15.0 km across a bay. (a) what was the avg speed in m/s? (b) At this speed, how much time would be required for the boat to take a 420 km trip up the river?
    THX
     
  2. jcsd
  3. Jun 17, 2004 #2
    For the bridge question, you know that the current velocity is given by [itex] v = v_i + at[/itex]. You know that [itex] v_i = 0, a = 9.8 \frac{m}{s^2}[/itex], and [itex] t = 2.50 s[/itex]. Plugging those numbers into the first equation, what number comes out?

    What about the distance it travelled? You know that you can relate distance, initial velocity, acceleration, and time with [itex] x = v_i t + \frac{1}{2}at^2[/itex], where x is the displacement. Using the numbers you already have from the previous part, what do you get for x? This is the distance the object has fallen - the same thing as the "height of the bridge".

    Average velocity is the ratio of displacement to time, [itex] v_{avg} = \frac{\Delta x}{\Delta t}[/itex]. You know the displacement from your recent work, and you know the time is 2.50 s, because that's given to you in the problem. What do you get when you divide the displacement by 2.50 s? That's the average velocity.

    For the ferryboat question, you can find the average velocity just like you did in the bridge question. Take the displacement (15.0 km) and divide it by the time (45.0 min). That's the average velocity. But it will be in kilometers per minute, and the problem wants you to give the average velocity in meters per second. No problem! Just multiply by a conversion factor so that the units you don't want go away, and the units you do want are all that's left. Remember, a conversion factor is equal to one, you can multiply anything you want by a conversion factor and you won't change its value.

    [tex] \frac {15.0 km}{45.0 min} \cdot \frac {1000 m}{1 km} \cdot \frac {1 min}{60 s} = 5.56 m/s[/tex]

    If you think of the units, like "km" and "min", as being just like variables, you can see that on the left hand side all the "km"s cancel out, as do all the "min"s, and you're just left with regular meters and seconds.

    To answer part (b), let's first get 420 km into the same units as the speed we know (meters):
    [tex] 420 km \cdot \frac {1000 m}{1 km} = 420000 m [/tex]

    So, what we're really asking is, how long does the boat need to go at 5.56 m/s before it has moved 420,000 m?

    We know that, with a constant velocity, [itex] x = vt[/itex], displacement is the product of velocity and time. If we solve that for t, we get:
    [tex] \frac {x}{v} = t [/tex]

    We know x (420,000 m) and v (5.56 m/s), so when we divide those out, what do we get?

    [tex] \frac {420000 m}{5.56 m/s} = 75600 s [/tex]

    It will take 75,600 seconds for the boat to go that distance. That's a pretty unwieldy number, though, so let's convert it into hours by dividing by 3600. 75,600 seconds is 21 hours.
     
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