Answer
The two planes passed each other 5.2 h after start of the flight and 5980 km away from Beijing airport (3517 km away from San Francisco airport)
Work Step by Step
Please see the attached image first.
Let's apply equation $S=Ut+\frac{1}{2}at^{2}$ for both planes seperately as follows.
For the Beijing plane $=\gt$
$\rightarrow S=Ut+\frac{1}{2}at^{2}$
Now plug the known values into this equation.
$S1=1150\space km/h\times t+\frac{1}{2}(0)t^{2}$
$S1 = 1150t\space km/h-(1)$
For the San Francisco plane $=\gt$
$\rightarrow S=Ut+\frac{1}{2}at^{2}$
Now plug the known values into this equation.
$S1=687\space km/h\times t+\frac{1}{2}(0)t^{2}$
$S1 = 687t\space km/h-(2)$
We know that ,$S1+S2= 9497\space km$
From (1),(2) $=\gt$
$1150t\space km/h+687t\space km/h=9497\space km$
$1837t=9497\space h$
$t=5.2\space h$ (amount of time to pass each other)
Let's apply this value to equation (1)
$S1=1150\space km/h\times 5.2\space h$
$S1=5980\space km$ (The two planes passed each other 5980 km away from Beijing airport)
