P01S2019 – Missiles Colliding

We set $x=0$ at the initial location of Spaceship A which therefore makes $x=D$ the initial location of Spaceship B.

The position $x_A\left(t\right)$ of the missile from Spaceship A is given by

x_A\left(t\right)=v_At

Since the missile from Spaceship B is launched at $t_0=2.0\ s$ from $x=D$, we conclude that its position $x_B\left(t\right)$ is given by

x_B\left(t\right)=-v_B\left(t-t_0\right)+D

We can then solve for the time $t_f$ that both missiles collide as follows

\begin{aligned}
x_A\left(t_f\right)=x_B\left(t_f\right)\ \ \ \ &\Rightarrow \ \ \ \ \ v_At_f=-v_B\left(t_f-t_0\right)+D \\
\\
&\Rightarrow \ \ \ \ \ v_At_f=-v_Bt_f+v_Bt_0+D \\
\\
&\Rightarrow \ \ \ \ \ \left(v_A+v_B\right)t_f=v_Bt_0+D \\
\\
&\Rightarrow \ \ \ \ \ t_f=\frac{v_Bt_0+D}{v_A+v_B} \\
\\
&\Rightarrow \ \ \ \ \ t_f=\frac{5\times {10}^3\cdot 2+990\times {10}^3}{3\times {10}^3+5\times {10}^3}=125\ s
\end{aligned}

To determine the positive where the two missiles collide, we can compute $x_A\left(t_f\right)$ or $x_B\left(t_f\right)$ (since they are equal) and we find that it is equal to

\boxed{x_A\left(t_f\right)=v_At_f=3\times {10}^3\cdot 125=375,000\ m=375\ km}