Solve this:

9. A particle moves along a path ABCD as shown in the figure. Then the magnitude of net displacement of the particle from position A to D is : 




1   10   m               2   5 2   m             3   9   m                 4   7 2 m

Dear student


Construction: Extend AB to F  and join DF such that  BC=EF=CE=BF=4m.Now in CEDby pythagoras theoremCD2=DE2+CE252=DE2+42DE=3 mSo, DF=DE+EF=3m+4m=7mand AF=AB+BF=3m+4m=7mSo displacement=AF2+FD2=72+72=49+49=98=72m
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