Speed of object towards a point not in the object's trajectory? Trying to study for my mid-term, but I'm having slight difficulties understanding what I'm supposed to do in this one problem:
A batter starts running towards first base at a constant speed of 6 m/s. The distance between each adjacent plate is 27.5 m. After running for 20 m, how fast is he approaching second base? At the same moment, how fast is he running away from third base? (see image below)

This is what I have so far:


*

*Let $d$ be the distance the batter has run thus far

*The distance between the batter and first base is 7.5 m

*The distance between the batter and second base is $\sqrt {27.5^2 + (27.5-d)^2}\ $, or approx. 28.5044 m when $d = 20$

*The distance between the batter and third base is $\sqrt {27.5^2 + d^2}\ $, or approx. 34.0037 m when $d = 20$


No need to hand feed me the answer, I'd just like a bit of insight on how to solve the problem.
 A: You don't need derivatives.  Just find the coordinates of the batter and the two bases of interest.  Now project the batter's velocity on the vector from him to each base.
By the way, one should not measure baseball fields in meters, but in feet.
A: For each case, draw a line from the runner to the relevant base, creating a right-angle triangle.  You know two of the sides of this triangle, so you know the third side and all trig ratios of all the angles.
Break the runners known speed into two components, one along the line drawn above, and the other  perpendicular to it. Find the angles in the components that match those from the triangle and use the trig ratios to evaluate the components.
A: The first question is easy: the batter is running straight towards the first base, so he is approaching the first base with a speed of $6$ m/s. To answer the second question, try to find the function $f(t)$ of time that gives the distance to the third base. Then find the derivative at the point $t=\frac{20}{6}$, when the batter has been running for 20 metres.
