How to add acceleration with velocity?

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.

 

Same as the big ship.

 

No. It will be moving faster, but accelerating at the same rate.

 

Yes.

 

No. Same acceleration does not imply same velocity.

 

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.

 

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

 

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.

 

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.

 

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).

 

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).