Answer to Question #5198 in Mechanics | Relativity for Brittany

Question

Two spacecraft are 13,500 m apart and moving directly toward each other. The first spacecraft has velocity 525 m/s and accelerates at a constant −15.5 m/s2. They want to dock, which means they have to arrive at the same position at the same time with zero velocity. What would the average acceleration of the second spacecraft be in m/s2? Hint: Use the first spacecraft to get needed information for the second spacecraft remembering that the time is the same for both and their total distance is 13,500 m. Assume the second spacecraft to have a positive initial velocity. Include a sign with your answer.

Answer 1

Let V1, V2, a1, a2 - velocities and accelerations of the first and the second
spacecrafts respectively.
Let x be the position of their docking and t is
time needed to arrive to the x.
Let the first spacecraft started from the
position 0 and let X-axis be directed from the first spacecraft to the
second.
Then we have that the second spacecraft starts from the position
13500.
Let's write a system of equations for displacements and velocities, as
we know that their final velocities must be zeroes:

x = V1*t -
(a1*t^2)/2,
x = 13500 - V2*t + (a2*t^2)/2,
V1 - a1*t = 0,
V2 - a2*t =
0.

From the third and the 4th equations we get

t = V1/a1,
V2 =
V1 * a2/a1.

When we substract the second equation from the first, we
get:

0 = V1*t - (a1*t^2)/2 - 13500 + V2*t - (a2*t^2)/2,
=>

0 = V1*V1/a1 - (a1*(V1/a1)^2)/2 - 13500 + V1 * a2/a1*V1/a1 -
(a2*(V1/a1)^2)/2,

0 = V1^2/(2*a1) - 13500 + (a2*(V1/a1)^2)/2,

0 =
a1 - 27000*(a1/V1)^2 + a2,

a2 = 27000*(a1/V1)^2 - a1,

a2 =
27000*(15.5/525)^2 - 15.5 = 8,

V2 = V1 * a2/a1 = 525*8/15.5 =
271.

So, the second spacecraft has initial velocity 271 m/s directed
toward the first spacecraft and acceleration -8 m/s^2.