Analyzing the Trajectory and Acceleration of a Rocket Launch.

In summary, the body will not return to its origin at time t = 2T as it will continue moving in the negative direction for an additional time √2T. The displacement vectors with respect to time can be formulated as 1/2(ax,ay,az)T^2 and -1/2(ax,ay,az)t^2, but the formula becomes more complex when considering the change in direction of acceleration. Additionally, the trajectory vector for this motion would require following different rules for different time intervals.
  • #1
peripatein
880
0
Hello,

Granted that a body begins accelerating from rest with a constant acceleration a, and after a certain time T changes the direction of its acceleration and begins moving with acceleration -a, would it be correct to say that the body would return to its origin at time t=2T? I mean, are the displacement vectors with respect to time merely 1/2(ax,ay,az)T^2 and -1/2(ax,ay,az)t^2?
 
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  • #2
Say you accelerate at 10 for 1 second - your speed would be 10.
If you then accelerate at -10 for 1 second, you speed would be zero. The distance from the start would still be increasing during this second stage since the velocity is still positive.
 
  • #3
at time t = T the body will have some velocity v = aT, whereas at time t = 0, the body has a velocity = 0

so if the body switches instantly from a to -a at time t = T, then it will first have to reduce its velocity to zero before it can start moving in the negative direction

so no, it will not return to its origin at t = 2T
 
  • #4
peripatein said:
Hello,

Granted that a body begins accelerating from rest with a constant acceleration a, and after a certain time T changes the direction of its acceleration and begins moving with acceleration -a, would it be correct to say that the body would return to its origin at time t=2T? I mean, are the displacement vectors with respect to time merely 1/2(ax,ay,az)T^2 and -1/2(ax,ay,az)t^2?

I think you are alluding here to a formula similar to

s = ut + 1/2 at2

For the first time interval "ut" is 0, so your 1/2(ax,ay,az)T^2 is appropriate
For the second time interval, "ut" is certainly NOT zero, so just -1/2(ax,ay,az)t^2 is insufficient.
 
  • #5
Would it then be correct to formulate its displacement vector thus:
r(t) = 1/2(ax,ay,az)T^2 + (-ax,-ay,-az)Tt + 1/2(-ax,-ay,-az)t^2?
 
  • #6
peripatein said:
Would it then be correct to formulate its displacement vector thus:
r(t) = 1/2(ax,ay,az)T^2 + (-ax,-ay,-az)Tt + 1/2(-ax,-ay,-az)t^2?

I don't agree with the negative signs in the red section.

The initial velocity for the second segment is the final velocity for the first segment, and in the first segment the accelerations were positive.
 
  • #7
Is its initial velocity in the second segment not in the opposite direction, as its new acceleration?
 
  • #8
peripatein said:
Is its initial velocity in the second segment not in the opposite direction, as its new acceleration?

yes, but you have negative signs on both the initial velocity AND the acceleration. IF they are in opposite directions, one of them has to be positive [and the acceleration definitely has the negative signs in your nomenclature].
 
  • #9
Okay, but then I get that its displacement would be equal to zero at t= (1+sqrt(2))*T.
Could it be? How come the solution for r0, i.e. when body is at the origin, does not include t0=0?
 
  • #10
I mean, what would be the formulation of its trajectory vector, if in the first segment it is 1/2at^2, whereas in the second segment it is 1/2aT^2+aTt-1/2at^2?
Is there any way to combine them into one vector?
 
  • #11
peripatein said:
I mean, what would be the formulation of its trajectory vector, if in the first segment it is 1/2at^2, whereas in the second segment it is 1/2aT^2+aTt-1/2at^2?
Is there any way to combine them into one vector?

This problem could mirror the following:

A rocket is launched and has an upward acceleration of 9.8 ms-2 while the engine is firing.
Once the engine runs out of fuel, the rocket is (naturally) subject to a downward acceleration of 9.8 ms-2.

If we were to derive an equation for the displacement, it would be a complex function where you follow one rule for certain domains, and a different rule for other domains.

ie:
0 --> T follow 4.9t2
T --> X follow 4.9T2 + 9.8T - 4.9t2

X is the time the rocket hits the ground again.

Next post will be a conceptual analysis of this.
 
  • #12
peripatein said:
I mean, what would be the formulation of its trajectory vector, if in the first segment it is 1/2at^2, whereas in the second segment it is 1/2aT^2+aTt-1/2at^2?
Is there any way to combine them into one vector?

Conceptual approach to rocket launch.

While the rocket motor is burning, the speed increases to a maximum (V). the distance covered (height gained) equals the average speed multiplied by the time interval (V/2 x t)
Once the rocket engine fails, the rocket slows. With the same numeric value of acceleration it will take the same time to stop. Its initial and final speeds will be the same as when it was accelerating up [just V initial and 0 final, rather than 0 initial and V final) so the extra distance traveled up will be an equal amount.

If the rocket gained height H with the engine burning, it will eventually reach maximum height of 2H

The s = ut + 1/at2 formula shows us that when u is zero, and s is doubled, t is increased by a factor of √2.
When this rocket, having reached a height of 2H, falls back to Earth this is indeed the situation.
So the Rocket will have traveled up, with engine firing, for time T.
It will have continued to drift up for an additional time T.
It will then fall back to Earth in time √2.T

Total trip: (2 + √2)T

or as you found (1 + √2)T from the end of the first segment of the motion.
 

What is trajectory?

Trajectory refers to the path that an object follows through space. It is typically described in terms of position, velocity, and acceleration.

What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity, meaning it has both magnitude and direction.

How is acceleration related to trajectory?

Acceleration plays a crucial role in determining an object's trajectory. The direction and magnitude of an object's acceleration will affect its path through space.

What factors affect trajectory and acceleration?

The factors that affect trajectory and acceleration include the object's mass, the force acting on the object, and any external forces such as air resistance or friction.

How can trajectory and acceleration be calculated?

Trajectory and acceleration can be calculated using equations such as the kinematic equations or Newton's laws of motion. These equations take into account the initial position, velocity, and acceleration of the object to determine its path through space.

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