Acceleration=10 NOT 9.8

Question

Two spacecraft are 23,500 m apart and moving directly toward each

other. Spacecraft#1 has velocity 535 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. These aliens are watching in anticipation. (a) What should the initial velocity of

the spacecraft#2 be? (b) What should be #2 constant acceleration? (Hint start with spacecraft

#1)

Answer 1

Hi Dessie C.,

You need to apply your basic equations of motion:

v(f)^2 = v(i)^2 + 2 a*d

and d = v*t + (1/2)a*t^2

as applicable. You should memorize these two equations, they're SO useful!

First, make a frame of reference, let's say based (zero position) where S#1 starts from. Its motion will therefore be (+) signed, and its acceleration (to come to a stop) (-) signed.

Calculate where S#1 ends up, using the first equation above. Then just divide its initial velocity by its acceleration to find out the time interval in question.

Next, calculate the distance S#2 must travel: the initial separation minus the distance S#1 travels.

Then, apply the first equation above to S#2. Remember that v(i)₂ will be (-) valued and a₂ will be (+) valued. You now have a relationship among v(i)₂^2 and a₂ .(but, neither yet solved for!)

Next, notice that you already do know what relates v(i)₂ and a₂ : v = a*t (right? That's another relationship you absolutely should memorize)

So substitute in a₂t for v(i)₂ and look at the form of the resulting equation: it's a quadratic in a₂ . Solve for a₂ , then substitute that back in to find v(i)₂ .

-- Happy spacings, -- Mr. d.