Astronaut A1= 65 Kg Astronaut A2= 72 Kg. They both agree to rendezvous in space
ID: 1961793 • Letter: A
Question
Astronaut A1= 65 Kg Astronaut A2= 72 Kg. They both agree to rendezvous in space after a few hours. They plan to let gravity bring them together, and they are 12 meters apart.A1 acceleration=3.2x10^-11
A2 acceleration=2.9x10^-11
(A) Make free body diagram....done
(B) If their acceleration remained constant, how many days would they have to wait before reaching each other?=9.4 days
(C) Irrelevant
(D) Measured from the position of A1, where do the astronauts meet?
(E) What are the astronauts speed at impact?
Explanation / Answer
Let A1 be at x=0m at t=0s, and A2 be at x = 12m t =0s.
the the equation for A1 is
x1(t) = (1/2) * 3.2 * (10^-11) * t^2
and for A2
x2(t) = 12 - (1/2) * 2.9* (10^-11) * t^2
B) I think you should check your answer for this part. When I solved for the distance travelled by each astronaut using your t, the total does not give me 12 meters. The equation for position is
x(t) = x_0 + v*t + (1/2) a * t^2
To solve B), you want the time when the position of the first particle equals the second particle (astronaut). Therefore, you want to set the two equations equal to each other, and solve for t. Note here that x_0 is the initial start position on the x-axis, and a will be positive for one astronaut and negative for the other.
Alternatively, you can just write an equation for the total distance travelled by both astronauts and set that equal to 12 and solve for t. You should arrive at the same answer.
D) Just plug in your t from part B into either equation for position, and you will find the point where they meet.
E) Speed is related to velocity by v(t) = v_initial + a*t. Again, t here is when they meet, so is the answer you arrive at in B. B is really the most difficult part of this problem. (I assume initial velocity is zero)
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