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I'm not sure what to do. Do I break the thing into shapes and calculate each?

Like:

$A = 2\times 4\times 0.5 = 4$
$B = 8\times 4 = 32$
$C = ?$
$D = ?$

Also, how does the $14s$ impact this?

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    Yes, a piecewise calculation will work. The question is asking you to calculate up to the 14-second mark. – 2017-02-22
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    So are A and B correct? How would I calculate the next two sections? – 2017-02-22
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    Also, would I just add A thru D together? – 2017-02-22
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    Your calculation for A and B gives the correct number of metres travelled in each of those periods, yes. I'm sure you can manage the other calculations. – 2017-02-22
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    But C is not a triangle? How would that work? – 2017-02-22
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    @user416503 $C$ is a right trapezoid. Just as easy to calculate its area as for a triangle. – 2017-02-22
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    So if I add everything, then answer is 48? Is there anything else htat must be done? – 2017-02-22

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Hint: when the speed varies linearly with time, the distance covered is the same as if the speed was constant and equal to the average of the starting and ending speeds. Therefore $C = 2 \cdot (4+2)/2$.

(In fact you already used this for $A = 2 \cdot 4 \cdot 0.5 = 4$ since $2$ is the time, and $4 \cdot 0.5 = \frac{0 + 4}{2}\,$ is the average between speed $0$ at the beginning of time interval $[0,2]$ and speed $4$ at its end.)

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    So if I add all, then the answer is 48? – 2017-02-22
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    No. How did you get $D\,$? Don't forget that the clock stops at $14\,$s. – 2017-02-22
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    I have A = 4; B = 32; I split C and D into seprate things - C, the triangle, is 2(4) (1/2)...D, the rectangle left over, is 4*2... Adding everything is 48. How would you solve it, if this is incorrect? – 2017-02-22
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    `C, the triangle, is 2(4) (1/2)` Sorry, I don't follow this part, and it's wrong. You can see the right $C$ in my answer above, and all that's left to do is add $D = 2 s \times 2 m/s\,$. – 2017-02-22