I will set these up for you and you do the math.

(1)

a = (v_{2} - v_{1})/t

v_{2} = 60 mi/hr

v_{1} = 30 mi/hr

t = 10 sec = 10/3600 hrs

Decide whether you want the acceleration in mi/s^{2} or mi/hr^{2}, and plug in the numbers to get a.

From kinematics,

v_{2}^{2} - v_{1}^{2} = 2ad ⇒

d = (v_{2}^{2} - v_{1}^{2}) / (2a)

Plug in the numbers in appropriate units and get the distance d.

(2)

Find how long it takes spacecraft 1 to stop, and how far it has traveled.

d = initial separation distance = 13500 m

v_{1}(t) = v_{1}_{i} + a_{1}t = 525 - 15.5t

Set v_{1}(t) = 0 and solve for t.

525 - 15.5t = 0

Solve for t. That is how long it takes to come to instantaneous rest. Its travel distance is

x_{1} = v_{1i}t + (1/2)a_{1}t^{2} = 525t - (15.5/2)t^{2}

Plug in the travel time t computed above, and get the travel distance x_{1}. Now use these results to calculate initial speed and acceleration of spacecraft 2. It travels for the same length of time as 1, so use the same t you calculated above.

v_{2}(t) = v_{2i} + a_{2}t

Set v_{2}(t) = 0

(i) v_{2i} + a_{2}t = 0

You know t, but you have 2 unknowns v_{2i} and a_{2}, so you need a second equation. The distance that 2 travels gives you the second equation.

x_{2} = d - x_{1}

x_{1} = distance traveled by 1, which you calculated above, and you know d.

(ii) d - x_{1} = v_{2}_{i}t + (1/2)a_{2}t^{2}

Plug d, x_{1}, and t into equations (i) and (ii) and solve for the unknwons v_{2}_{i} and a_{2}. You have a system of 2 linear equations in 2 unknowns, which you should know how to solve. There will be a lot of number crunching!