$A$ frog hops along a straight line path from point $A$ to point $B$ in $10 \, s$ and then turns and hops to point $C$ in another $5 \, s$. Based on the provided number line,calculate the average speed and average velocity of the frog for the motion between: $(a)$ $A$ to $B$ $(b)$ $A$ to $C$ (through $B$).
(N/A) From the number line,the positions are: $A = 7 \, m$,$B = -2 \, m$,$C = 3 \, m$.
$(a)$ For motion from $A$ to $B$:
Total distance $= |(-2) - 7| = 9 \, m$.
Total displacement $= -2 - 7 = -9 \, m$.
Time taken $= 10 \, s$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{9}{10} = 0.9 \, m/s$.
Average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{-9}{10} = -0.9 \, m/s$.
$(b)$ For motion from $A$ to $C$ through $B$:
Total distance $= |(-2) - 7| + |3 - (-2)| = 9 + 5 = 14 \, m$.
Total displacement $= 3 - 7 = -4 \, m$.
Total time $= 10 + 5 = 15 \, s$.
Average speed $= \frac{14}{15} \approx 0.933 \, m/s$.
Average velocity $= \frac{-4}{15} \approx -0.267 \, m/s$.
$(a)$ For motion from $A$ to $B$:
Total distance $= |(-2) - 7| = 9 \, m$.
Total displacement $= -2 - 7 = -9 \, m$.
Time taken $= 10 \, s$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{9}{10} = 0.9 \, m/s$.
Average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{-9}{10} = -0.9 \, m/s$.
$(b)$ For motion from $A$ to $C$ through $B$:
Total distance $= |(-2) - 7| + |3 - (-2)| = 9 + 5 = 14 \, m$.
Total displacement $= 3 - 7 = -4 \, m$.
Total time $= 10 + 5 = 15 \, s$.
Average speed $= \frac{14}{15} \approx 0.933 \, m/s$.
Average velocity $= \frac{-4}{15} \approx -0.267 \, m/s$.
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