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How to add acceleration with velocity?

  1. Oct 1, 2013 #1
    Suppose I'm in a rocked ship and its accelerating away from earth at a speed of 9m/s^2. From inside my rocket i start another toy rocket which moves in a uniform velocity from my reference frame at 1m/s in the direction of motion of my ship (Please note the direction of motion of both ship is same)
    But for an observer on earth the toy ship isn't moving at a constant velocity, it is accelerating. What will be the rate of acceleration of the toy ship for him?
     
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  3. Oct 1, 2013 #2

    UltrafastPED

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    The second ship has an initial velocity equal to the velocity of the first ship from the instant it was launched, v_1(t), and now your second ship is given a uniform velocity v_2; the velocity as seen by the stationary observer (at the original launch site) is v_1(t) + v_2.

    Since the first ship has a constant acceleration of 9 m/s^2, and started from rest at the launch site we can obtain v_1(t) = 0.5 x 9 t^2 (Galileo's formula) = 4.5 t^2.

    Since v_2 was given as 1 m/s we can now say that the velocity of the second ship, the toy, is
    V_toy = 1 + 4.5 t^2 [m/s], where t is the time when the toy ship was launched.

    The toy ship is _not_ accelerating after it has been launched - you only specified that the first ship was accelerating - presumably due to the firing of its rockets, which the toy ship lacks.
     
  4. Oct 1, 2013 #3

    A.T.

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    Same as the big ship.
     
  5. Oct 1, 2013 #4
    its not as same as the big ship, for an observer on ground the toy ship will be accelerating faster than the mother ship
     
  6. Oct 1, 2013 #5

    A.T.

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    No. It will be moving faster, but accelerating at the same rate.
     
  7. Oct 1, 2013 #6
    the toy ship is in an uniform velocity with respect to the the mother ship, but the toy ship is accelerating with respect to the ground, am i right?
     
  8. Oct 1, 2013 #7
    how can you say the toy ship is accelerating at same rate?
    what if the toy ship was moving with a greater velocity about 10m/s ?
    If both wet accelerating at same rate then the relative velocity between the two ships must be 0
     
  9. Oct 1, 2013 #8

    A.T.

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    Yes.
     
  10. Oct 1, 2013 #9

    A.T.

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    No. Same acceleration does not imply same velocity.
     
  11. Oct 1, 2013 #10

    UltrafastPED

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    From the conditions which you have given the toy ship is _not_ accelerating after it has been launched. You said that the "rocket ship" is accelerating at 9 m/s^2 ... this means that it is the rocket which is providing its acceleration.

    Thus the toy ship is being accelerated at the same rate as the rocket ship up until it is launched: after that it sees no acceleration. Please look at the calculations which I provided - they fit the conditions which you stated exactly, and provide a complete answer.
     
  12. Oct 1, 2013 #11

    A.T.

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    Just to clarify:
    Thats not how I interpret the OP. My answers assume constant velocity of small rocket, relative to big rocket after launch.
     
  13. Oct 1, 2013 #12

    UltrafastPED

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    But that is inconsistent with "toy launched at 1 m/s" vs the big ship has rockets.

    Plus the OP has a lot of confused statements ... lets request that he clarify.

    If the toy does have its own rocket, and accelerates at 9 m/s^2 then its velocity will be 1 m/s greater than the big rocket due to its initial velocity of 1 m/s. In that case both ships would have the same acceleration, but would have different velocities.
     
  14. Oct 1, 2013 #13
    Trojan, are you looking for a relativistically correct formula? AT's answer is correct for nonrelativistic velocities. For relativistic velocities, I believe the coordinate acceleration of the toy rocket (using Earth coordinates) must be less, especially if the toy rocket is traveling near c in the Earth frame. The proper acceleration is the same for both rockets.

    If your big rocket has constant 9 m/s^2 proper acceleration, the acceleration as viewed by Earth will be decreasing in time as your rocket gets closer to c.

    The number you gave, 1 m/s, is very nonrelativistic, so just use AT's answer.
     
    Last edited: Oct 1, 2013
  15. Oct 1, 2013 #14
    It does not say anything about "launching".
    It says: "another toy rocket which moves in a uniform velocity from my reference frame"
    Which is compatible with A.T.'s interpretation.
    Uniform velocity refers to the duration of the motion and not to an instant (like the launching time).
     
  16. Oct 1, 2013 #15
    we are not moving any closer to c, that will be complicated.
    Just for last time I'm asking it again
    Ill call the Big ship in which I'm sitting as "A" and the toy ship as "B"
    Now just imagine everything from the observer on earth’s frame of reference. The observer on earth sees the rocket "A" moving at an acceleration of 9m/s2, after some time (I'm not giving any importance to a specific time ) the toy ship that is "B" launched with a constant velocity wrt to the A ship. So the observer on ground sees the toy ship "B" moving away from the big ship "A", am i right?
    Well "MOVING AWAY" WHICH MEANS IT HAS GREATER SPEED THAN THAT OF BIG ROCKET"
    Then how can the observer on Ground say the toy ship is in a constant velocity? or same acceleration as 9m/s2?
    Because it was already accelerating along with the big ship, now it got additional thrust so it will be moving faster than big rocket, am i right?
     
    Last edited: Oct 1, 2013
  17. Oct 1, 2013 #16

    A.T.

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    During the launch, when the toy gets that 1m/s extra speed, it has temporally a higher acceleration than the big ship. But once it moves at constant speed relative to the big ship, it has the same acceleration as the big ship in any frame (ignoring relativity).
     
    Last edited: Oct 2, 2013
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