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Question Number 29268 by tawa tawa last updated on 06/Feb/18

A man moves  20m  North ,  then  12m East  and  finally  15m  South.His  displacement from the starting point is now  (a) 13m  (b)  27m  (c)  47m  (d)  23m

$$\mathrm{A}\:\mathrm{man}\:\mathrm{moves}\:\:\mathrm{20m}\:\:\mathrm{North}\:,\:\:\mathrm{then}\:\:\mathrm{12m}\:\mathrm{East}\:\:\mathrm{and}\:\:\mathrm{finally}\:\:\mathrm{15m}\:\:\mathrm{South}.\mathrm{His} \\ $$$$\mathrm{displacement}\:\mathrm{from}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{point}\:\mathrm{is}\:\mathrm{now} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{13m}\:\:\left(\mathrm{b}\right)\:\:\mathrm{27m}\:\:\left(\mathrm{c}\right)\:\:\mathrm{47m}\:\:\left(\mathrm{d}\right)\:\:\mathrm{23m} \\ $$

Commented by Rasheed.Sindhi last updated on 06/Feb/18

(a)13 m  Net distance along north 20−15=5 m  Distance to the east 12  Displacement (√(5^2 +12^2 ))=(√(169))=13 m

$$\left(\mathrm{a}\right)\mathrm{13}\:\mathrm{m} \\ $$$$\mathrm{Net}\:\mathrm{distance}\:\mathrm{along}\:\mathrm{north}\:\mathrm{20}−\mathrm{15}=\mathrm{5}\:\mathrm{m} \\ $$$$\mathrm{Distance}\:\mathrm{to}\:\mathrm{the}\:\mathrm{east}\:\mathrm{12} \\ $$$$\mathrm{Displacement}\:\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\sqrt{\mathrm{169}}=\mathrm{13}\:\mathrm{m} \\ $$

Commented by tawa tawa last updated on 06/Feb/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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