1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Six more kinematics, worked out already with questions/doubts/wrong answers

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    1. A geosynchronous satellite travels around the earth once every 24 hours (thereby always staying above the same point on the earth's surface). Such satellites are at a distance of 4.23*10^7 m from the center of teh earth. How fast is such a satellite moving with respect to the earth?

    Answer: 3.08 * 10^3 m/s

    2. A long distance swimmer is able tos wim through still water at 4 km/h. She wishes to try to swim from Port Angeles, WA due north to Victoria BC, a distance of 50 km. An ocean current flows through the Strait of Juan de Fuca from west to east at 3 km/h. In what direction should she swim to make the crossing along a straight line between the two cities?

    a. 37 degrees west of north
    b. 37 degrees east of north
    c. 41 degrees west of north
    d. 41 degrees east of north
    e. 49 degrees west of north

    Answer: e

    3. The x and y coordinates of a particle in motion as functions of time, t, are given by:
    [tex]x = 3t^{2} - 12t + 2[/tex]
    [tex]y = 2t^{3} - 3t^{2} - 12t - 4[/tex]

    The smallest magnitude of the acceleration is closest to:

    a. 0
    b. 6.0 m/s^2
    c 8.5 m/s^2
    d. 12 m/s^2
    e. 13.4 m/s^2

    Answer: b

    4. Two bullets are fired simultaneously uphill parallel to an inclined plane. The bullets have different masses and different initial velocities. Which will strike the plane first?

    a. the fastest one
    b. the slowest one
    c. the heaviest one
    d. the lightest one
    e. they strike the plane at the same time

    Answer: e

    5. A helicopter is approaching the deck of the ship on which the pilot wishes to land. The ship is moving at 2.9 m/s west with respect to the water, which is flowing 8.9 m/s east. The helicopter is moving at 14m/s west with respect to the air, and the wind is blowing 15 m/s east with respect to the ground. What is the speed and direction of the ship's motion relative to the helicopter?

    Answer: 5 m/s east

    6. The compass of an aircraft indicates that it is headed due east, and its airspeed indicator shows that it is moving through the air at 150 km/hr. After flying for 2 hours, the aircraft is 350 km east and 74 km south of its starting point. What is the magnitude and direction of the wind velocity?

    Answer: 45 km/hr @ 55.95 degreees

    2. Relevant equations


    3. The attempt at a solution

    1. I worked this problem out and received the right answer, but I have a specific question for it. Here is the work:

    [tex]v = \frac{2\pir}{T} = \frac{2\pi(4.23 x 10^{7}}{24 x 3600} = 3076.15[/tex]

    However, I was wondering, before I checked the answer, whether I ought to have taken the speed of revolution of the Earth into account or not? Obviously, by the answer, I shouldn't have, but I'd like to know why? What does it mean when it asks for a velocity with respect to something, in this case, Earth?

    2. My vector diagram contains a vertical vector going north, which would be the straight line path. Above that is the vector pointing to the east, which is the current that flows 3 km/hr, and then the hypotenuse of this triangle is the 4 k/mhr vector of the swimmer. To calculate the angle between the vertical component and the hypotenuse, I used arc sin (3/4) to get a value of 48.5 degrees, which is the correct value, however, in my drawing, I think it ought to be east of north. You can see it in attachment #2.

    3. I am not sure how to determine the smallest magnitude of the acceleration. Could someone provide a starting point?

    4. I am not quite good with conceptual questions. Could someone help me understand why they will strike at the same time?

    5. For this one I determined the speed of the ship by subtracting 8.9 m/s, the flowing water, from 2.9, the speed of the ship, to get 6. I then did the same with the helicopter, 15-14, to get 1. Both are in the same direction. I'm not sure how I should get the speed of the ship relative to the helicopter, whether to add these two values or to subtract them. Obviously, by the answer, subtract, but I do not quiet understand why. I am a little fuzzy on the relative velocity concept, I'm afraid.

    6. This one I have absolutely no idea how to begin solving. Any tips?

    Attached Files:

    • #2.jpg
      File size:
      5 KB
  2. jcsd
  3. Aug 25, 2009 #2


    User Avatar
    Homework Helper

    3) Find the x component of the acceleration by finding d^2(x )/(dt)^2.
    Similarly find the y component of the acceleration.
    You can find that x component of the acceleration is independent of time whereas the y component of acceleration depends on time t. Hence the minimum acceleration is the x component.
    Last edited: Aug 26, 2009
  4. Aug 26, 2009 #3
    1) Good point. Question should have stated "with respect to centre of the earth" to be more precise. Clearly in this case you've done what they wanted

    2) Exactly the same thing as your other thread. Pay attention to the directions in your vector diagram! You have to swim against the flow to counter it!

    3) See post #2

    4) Draw the problem out on paper, then tilt it such the incline is horizontal. Your initial velocities should now be horizontal too. Now, this is just a typical projectile motion question. Can you see the answer from this point onwards?

    5) If you hadn't already realized, what you did initially was to find, separately, the speed of the boat and helicopter relative to the GROUND. Now since they both have a common reference (the ground), you subtract them to get their relative speed, yes?

    6) Very similar to question 2. Plane flying east, wind blowing south, therefore...?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook