The following graph describes the motion of a girl going to meet her friend who stays $50 \ m$ from her house.
$(i)$ How much time does she take to reach her friend's house?
$(ii)$ What is the distance travelled by the girl during the time-interval $0$ to $12$ minutes?
$(iii)$ During which time-interval is she moving towards her house?
$(iv)$ For how many minutes was she at rest during the entire journey?
$(v)$ Calculate the speed by which she returned home.
$(i)$ How much time does she take to reach her friend's house?
$(ii)$ What is the distance travelled by the girl during the time-interval $0$ to $12$ minutes?
$(iii)$ During which time-interval is she moving towards her house?
$(iv)$ For how many minutes was she at rest during the entire journey?
$(v)$ Calculate the speed by which she returned home.
(N/A) $(i)$ Looking at the graph,the girl reaches the $50 \ m$ mark (friend's house) at $14$ minutes.
$(ii)$ From $0$ to $2$ min,distance $= 20 \ m$. From $2$ to $4$ min,distance $= 0 \ m$. From $4$ to $6$ min,distance $= 40 - 20 = 20 \ m$. From $6$ to $8$ min,distance $= 0 \ m$. From $8$ to $10$ min,distance $= 40 - 20 = 20 \ m$. From $10$ to $12$ min,distance $= 0 \ m$. Total distance $= 20 + 0 + 20 + 0 + 20 + 0 = 60 \ m$.
$(iii)$ She moves towards her house when the distance from the house decreases. This happens during the intervals $8$ to $10$ minutes and $14$ to $16$ minutes.
$(iv)$ She is at rest when the distance remains constant (horizontal line). This occurs during $2-4$ min ($2$ min),$6-8$ min ($2$ min),and $10-12$ min ($2$ min). Total time at rest $= 2 + 2 + 2 = 6$ minutes.
$(v)$ She returns home from $14$ to $16$ minutes. Distance covered $= 50 \ m$. Time taken $= 2$ minutes. Speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{50 \ m}{2 \ min} = 25 \ m/min$.
$(ii)$ From $0$ to $2$ min,distance $= 20 \ m$. From $2$ to $4$ min,distance $= 0 \ m$. From $4$ to $6$ min,distance $= 40 - 20 = 20 \ m$. From $6$ to $8$ min,distance $= 0 \ m$. From $8$ to $10$ min,distance $= 40 - 20 = 20 \ m$. From $10$ to $12$ min,distance $= 0 \ m$. Total distance $= 20 + 0 + 20 + 0 + 20 + 0 = 60 \ m$.
$(iii)$ She moves towards her house when the distance from the house decreases. This happens during the intervals $8$ to $10$ minutes and $14$ to $16$ minutes.
$(iv)$ She is at rest when the distance remains constant (horizontal line). This occurs during $2-4$ min ($2$ min),$6-8$ min ($2$ min),and $10-12$ min ($2$ min). Total time at rest $= 2 + 2 + 2 = 6$ minutes.
$(v)$ She returns home from $14$ to $16$ minutes. Distance covered $= 50 \ m$. Time taken $= 2$ minutes. Speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{50 \ m}{2 \ min} = 25 \ m/min$.
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