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Homework Help: Velocity and Time Functions HW

  1. Oct 10, 2005 #1
    I actually came up with some answers to these questions but I wanna be sure I have the *right* ones before I turn them in. If I have it right i'd appreciate the say-so if anyone minds the bother, but if i'm wrong maybe just some insight as to how to do better.

    A spelunker finds a keep hole in a cave. She drops a pebble into the hole and hears a splash two secs later. She remembers that the acc. due to gravity is about 10 m/s, and that sound travels at 300 m/s. From this, she determines the depth of the hole, taking into account that it takes time for the sound to reach her. How deep is the hole?

    Using the equation delta x = Vi deltaT + 1/2 a delta T squared, I came up with 20 meters for the depth of the whole, which made me feel pretty stupid as I could have deduced that just from seeing that it took a pebble two secs to make a sound at 10 m/s. However, I know that the sound and the time it takes to reach her ears must be accounted for as well, and I dont know how. I know any amount of time I get from the sound is going to be very tiny since it goes so fast, but I cant shake the feeling that i'm missing something. Anyone?


    A person sees a flowerpot falling past a window. The window is 1.5 m high, and it takes 0.3 secs for the pot to fall past the window.
    --Find speed that the flowerpot had when it was on top of the window.
    --Using result, find height from which the pot fell, assuming it fell from rest.

    For the first part, I got 4.9 m/s for the speed. For the second, I got that it fell from 1.91 meters. Right? Wrong? Horribly wrong?


    A car is at a constant speed of 55 m/s. It passes a motorcycle at rest by the side of the road. Just as the car passes, the cycle begins to accelerate with constant acceleration of 3.5 m/s.
    --Write an equation for the position of the car as a function of time. Write another equation for the position of the cycle as a function of time. Now write an equation that expresses the condition that the cyle caught up with the car. Use this to find the time that they cycle needs to catch the car.

    I actually just dont know where to start on this one. I dont know if I should make the equation start with delta T, or delta X, or whatever. Any help would be appreciated, and thanks in advance.
  2. jcsd
  3. Oct 10, 2005 #2


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    For problem one. Consider that the distance going down can be calculated as: [tex]d = \frac{1}{2}gt^2[/tex]. The distance coming up is [tex] d = v_s_o_u_n_d t_2 [/tex]. Fortunately the two d's are equal, and we have the value for t1 + t2.
  4. Oct 10, 2005 #3
    Hmm. Ok, after doing it again I got 18.81 meters for the depth of the hole. It took 0.06 seconds for the sound to get to her ear, so it actually took 1.94 seconds for the pebble to hit the water. Putting that into the equation I got 18.81 for the depth.
  5. Oct 10, 2005 #4
    I think that's a bit off as .06 in the D = Vsound * T equation will say it's 18 meters deep
  6. Oct 10, 2005 #5
    Aw, your right.
  7. Oct 10, 2005 #6
    Take it out to a few more decimal places and it should be good.

    For the third one since the car is moving at D(car)= (blah) and the cycle is moving at D(cycle)= (blah blah). Can you combine those equations to simply give something that says the cycle will catch up? Like D(cycle) = D(car) = etc?
  8. Oct 10, 2005 #7
    I honestly dont know. Can I?
  9. Oct 10, 2005 #8
    Ok. For #3, for the first equation I got deltaX = deltaT (55 m/s)

    For the second equation, I got deltaX = deltaT (3.5 m/s squared)

    For the equation expressing the condition that the cycle catches up with the car, I got deltaX = Vi deltaT + 1/2(3.5)deltaT (squared)

    I got that the cycle takes 31.97 seconds to catch the car, and that it catches the car 1,758.5 meters down the road.

    How'd I do?
  10. Oct 11, 2005 #9


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    The condition for the bike to catch up with the car is that the displacements are equal, that is [tex]d_b_i_k_e = d_c_a_r[/tex]. We know the equations for both. One can be described by the constant velocity equation and the other with constant acceleration. You will find that there will be two solutions when solving for time. That is when d = 0, and your REAL answer d = ?

    The check is pretty easy. Just plug the times (your answer) into both the separate equations for the bike and car, and see if you get the same distance. If you do then that means they are side by side (caught up condition met). ^^
  11. Oct 11, 2005 #10


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    For question #2. There are two parts to this question. Regarding the first part. You can imagine that there is only one initial velocity that will satisfy the condition of the flower pot being able to fly by 1.5 meters in only 0.3 seconds (given that acceleration = g is constant). Using the kinematics equation for constant acceleration:

    [tex] d = v_i t + \frac{1}{2}gt^2[/tex]
    We see that we are in fact solving for [tex]v_i[/tex] or its initial velocity as all other variables are known.

    Now the second part. The [tex]v_i[/tex] from the first part will now be known as [tex]v_f[/tex] or final velocity. We want to know how far something needs to drop to reach this velocity. Of course, we assume zero initial velocity and only gravitational acceleration. Another of your kinematics equations reads:

    [tex]v_f^2 = v_i^2 + 2ad[/tex]
    I think you can handle from here ;)
    Last edited: Oct 11, 2005
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