check the picture below.
now, hmmm I notice the came angle rate is positive,.... meaning the camera angle is increasing, no decreasing, therefore is really turning away from the goal line, meaning the runner is not going towards the goal line, but moving away from it, was wondering, if the runner is going to the goal line, the 0.4 rad/sec should be negative, but anyway, is not.
keeping in mind that, "x" distance of 5 yards, is a constant, whilst "y" is not, and neither is the angle.
[tex]\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{1}{5}y\implies \stackrel{chain-rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{5}\cdot \cfrac{dy}{dt}
\\\\\\
\cfrac{1}{cos^2(\theta )}\cdot \cfrac{d\theta }{dt}=\cfrac{\frac{dy}{dt}}{5}\implies \cfrac{5}{cos^2(\theta )}\cdot \cfrac{d\theta }{dt}=\cfrac{dy}{dt}\\\\
-------------------------------\\\\[/tex]
[tex]\bf tan(\theta )=\cfrac{10}{5}\implies tan(\theta )=2\implies \measuredangle \theta =tan^{-1}(2)\implies \measuredangle \theta \approx \stackrel{radians}{1.10715}
\\\\\\
\cfrac{d\theta }{dt}=\stackrel{rad/sec}{0.4}\\\\
-------------------------------\\\\
\cfrac{5\cdot 0.4}{cos^2(1.10715)}=\cfrac{dy}{dt}[/tex]
which ends up at 10.
make sure your calculator is in Radian mode, since the angle is in radians.
