Describing Motion Around Us: CBSE Class 9 Physics practice questions

D. 20 days

Together they close the gap by 7 + 5 = 12 yojanas per day, so they meet after $\dfrac{240}{12} = 20$ days (by then one has walked 140 yojanas and the other 100). 120 days uses the difference 7 − 5, which would fit only if they walked the same way; 48 and about 34 days use one messenger's speed alone. Speed as distance ÷ time was known in India at least from the Aryabhatiya (5th century CE).

B. $60\ \text{km h}^{-1}$ and $0\ \text{km h}^{-1}$

Average speed = $\dfrac{\text{total distance}}{\text{time interval}} = \dfrac{150 + 150}{2 + 3} = \dfrac{300}{5} = 60\ \text{km h}^{-1}$. The family ends where it started, so the displacement is zero and the average velocity is $0\ \text{km h}^{-1}$. 62.5 km/h comes from averaging the two leg speeds (75 and 50 km/h), which is wrong because the legs took different times.

D. 80 m

Displacement = shaded area = rectangle + triangle = $(4 \times 8) + \tfrac{1}{2} \times 8 \times (16 - 4) = 32 + 48 = 80\ \text{m}$. 128 m multiplies the final velocity by the time, as if the scooter had moved at 16 m/s throughout; 48 m and 32 m each leave out one part of the area.