Rasik walks 20 , m North. Then he turns right | vedclass

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(D) Let the starting point be $A$. $1$. Rasik walks $20 \, m$ North to point $B$. $2$. He turns right and walks $30 \, m$ East to point $C$. $3$. He t

Vedclass · Accepted Answer

$45 \, m$ East

Vedclass · Answer

(D) Let the starting point be $A$. $1$. Rasik walks $20 \, m$ North to point $B$. $2$. He turns right and walks $30 \, m$ East to point $C$. $3$. He turns right and walks $35 \, m$ South to point $D$. $4$. He turns left and walks $15 \, m$ East to point $E$. $5$. He turns left and walks $15 \, m$ North to point $F$. To find the distance from $A$ to $F$: Horizontal distance $= BC + DE = 30 \, m + 15 \, m = 45 \, m$. Vertical distance $= AB - CD = 20 \, m - 35 \, m = -15 \, m$ (meaning he is $15 \, m$ South of the horizontal line passing through $A$). However,looking at the path: Net horizontal displacement $= 30 \, m + 15 \, m = 45 \, m$ East. Net vertical displacement $= 20 \, m - 35 \, m + 15 \, m = 0 \, m$. Since the net vertical displacement is $0$,point $F$ is exactly $45 \, m$ East of point $A$.

Rasik walks $20 \, m$ North. Then he turns right and walks $30 \, m$. Then he turns right and walks $35 \, m$. Then he turns left and walks $15 \, m$. Then he again turns left and walks $15 \, m$. In which direction and how many metres away is he from his original position?

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