If [tex]d[/tex] is the distance the runner travels between first and second base, then the distance between him and home plate [tex]h[/tex] satisfies
[tex]h^2=d^2+90^2[/tex]
(where the 90 here refers to the distance between first base and home)
Differentiating both sides with respect to time [tex]t[/tex] gives
[tex]2h\dfrac{\mathrm dh}{\mathrm dt}=2d\dfrac{\mathrm dd}{\mathrm dt}\implies\dfrac{\mathrm dh}{\mathrm dt}=\dfrac dh\dfrac{\mathrm dd}{\mathrm dt}[/tex]
The runner's speed is [tex]\frac{\mathrm dd}{\mathrm dt}=25\frac{\rm ft}{\rm s}[/tex]. When he is [tex]d=30\,\mathrm ft[/tex] away from first base, he is
[tex]h=\sqrt{30^2+90^2}=30\sqrt{10}\,\mathrm{ft}[/tex]
away from home plate. So he is moving away from home plate at a rate of
[tex]\dfrac{\mathrm dh}{\mathrm dt}=\dfrac{30\,\rm ft}{30\sqrt{10}\,\rm ft}\left(25\dfrac{\rm ft}{\rm s}\right)=5\sqrt{\dfrac52}\dfrac{\rm ft}{\rm s}\approx7.906\dfrac{\rm ft}{\rm s}[/tex]
