Answer:
Total
displacement
=
sum of areas
\[(\Delta
ABF+rect.BCGF+\Delta CDE+rect.CEIG)\]
\[=\left[
\frac{1}{2}(AF\times FB)+(BC\times CG) \right.\]
\[\left.
+\frac{1}{2}(CE\times DJ)+(CG\times GI) \right]\]
\[=\left[ \frac{1}{2}(2\times 5)+(4\times
5)+\frac{1}{2}(4\times 5)+5(5\times 4) \right]=55\text{m}\]
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