SEE ATTACHMENT #1 for a diagram of initial conditions.
Spacecraft A is initially 120 m from spacecraft B, moving towards each other on parallel paths. Both spacecrafts apply reverse thrust to reverse their motion after which they meet twice. Spacecraft A has initial velocity VoA= +25 m/sec and acceleration aA= -3 m/sec^2. Spacecraft B has initial velocity VoB= -30 m/sec and acceleration aB=+ 4 m/sec^2. Let A be initially at the origin, and B at 120 m.
PART a. Find the two times and locations where they meet.
PART b. On the 'x vs t' set of axes show qualitatively, the x(t) graph of each spacecraft. Include x and t coordinates of intersections and intercepts.© BrainMass Inc. brainmass.com March 4, 2021, 5:43 pm ad1c9bdddf
See attached file.
PART a. Step 1.
The general motion equation, x(t), for a constant acceleration is:
(1) x = xo + Vo t + .5 a t^2
For A, therefore, we can write:
(2) xA = 0 + 25 t - 1.5 t^2
For B, ...
The two spacecraft approaches are analyzed to determine the time and locations. With a graph and clear explanations, the problems are solved.