Distance-function and velocity-function problem

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Matthewkind
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The problem is three paragraphs with a bunch of blank spots.

Starting from f(0) = 0 at constant velocity v, the distance function is f(t) = __[A]__. When f(t) = 55t the velocity is v = ____. When f(t) = 55t + 1000 the velocity is still __[C]__ and the starting value is f(0) = __[D]__. In each case v is the __[E]__ of the graph of f. When __[F]__ is negative, the graph of __[G]__ goes downward. In that case area in the v-graph counts as __[H]__.

Forward motion from f(0) = 0 to f(2) = 10 has v = ____. Then backward motion to f(4) = 0 has v = __[J]__. The distance function is f(t) = 5t for 0 <= t <= 2 and then f(t) - __[K]__.

_____________________________________________

There's more to the problem than this, but [K] is where I get stuck, not understand the problem. My answers thus far are:
[A] = vt, = 55, [C] = 55, [D] = 1000, [E] = slope, [F] = v, [G] = f, [H] = f, = 5, [J] = 0.

First, am I doing this correctly? And second, what precisely is K asking me for?
 
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Matthewkind said:
The problem is three paragraphs with a bunch of blank spots.

Starting from f(0) = 0 at constant velocity v, the distance function is f(t) = __[A]__. When f(t) = 55t the velocity is v = ____. When f(t) = 55t + 1000 the velocity is still __[C]__ and the starting value is f(0) = __[D]__. In each case v is the __[E]__ of the graph of f. When __[F]__ is negative, the graph of __[G]__ goes downward. In that case area in the v-graph counts as __[H]__.

Forward motion from f(0) = 0 to f(2) = 10 has v = ____. Then backward motion to f(4) = 0 has v = __[J]__. The distance function is f(t) = 5t for 0 <= t <= 2 and then f(t) - __[K]__.

_____________________________________________

There's more to the problem than this, but [K] is where I get stuck, not understand the problem. My answers thus far are:
[A] = vt, = 55, [C] = 55, [D] = 1000, [E] = slope, [F] = v, [G] = f, [H] = f, = 5, [J] = 0.

First, am I doing this correctly? And second, what precisely is K asking me for?

[J] is incorrect. The others are correct.