Two spacecraft are 13,500 m apart and moving directly toward each other. spacecraft#1 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. these aliens are watching in anticipation. (a) what should the initial velocity of the spacecraft#2 be? (b) what should be #2 constant acceleration?

-272 m/s, 8.03 m/s²

Explanation:

Let's call position 0 the point where the two spacecrafts meet.  Let's say spacecraft #1 starts x units to the left of 0 and spacecraft #2 starts 13500-x units right of 0.

For spacecraft #1:

v = 0 m/s

v₀ = 525 m/s

a = -15.5 m/s²

x = 0 m

x₀ = -x

The distance that spacecraft #1 travels is:

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (525 m/s)² + 2(-15.5 m/s²)(0 m − -x)

0 = 275625 − 31x

x ≈ 8891 m

The time it takes to reach 0 velocity is:

v = at + v₀

0 m/s = (-15.5 m/s²) t + 525 m/s

t ≈ 33.87 s

For spacecraft #2:

v = 0 m/s

x = 0 m

x₀ = 13500-x ≈ 4609 m

t = 33.87 s

Using the same equations:

v² = v₀² + 2a(x - x₀)

(0 m/s)² = v₀² + 2a(0 m − 4609 m)

0 = v₀² − 9218a

v = at + v₀

0 m/s = a(33.87 s) + v₀

0 = 33.87a + v₀

Now we have two equations and two variables.  We can solve the system of equations.

First, let's solve for a and substitute:

0 = 33.87a + v₀

a = -v₀ / 33.87

0 = v₀² − 9218 (-v₀ / 33.87)

0 = v₀² + 272 v₀

0 = v₀ (v₀ + 272)

v₀ = -272 m/s

Now we can find the acceleration:

a = -(-272) / 33.87

a = 8.03 m/s²

heats surface water up

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form