Motion in a Straight Line: NEET previous year questions

Throughout the flight the only acceleration is g directed downward, so $dv/dt$ stays a constant negative value. The velocity therefore decreases linearly, reaches zero at the highest point and turns negative on the way down, tracing a single straight line of constant negative slope, which is plot C. Hence option (1). Plots with a change of slope, a V-shape, or a positive slope contradict the constant downward acceleration.

The ruler falls freely, so distance $s = \frac{1}{2}gt^2$ increases monotonically with reaction time. Ranking the reaction times from largest to smallest gives B(0.22) > E(0.21) > A(0.20) > D(0.19) > C(0.18), which is the same order as the distances. Option (4) is the exact reverse (smallest first). Option (2) is also smallest first and swaps persons B and E as well. Option (1) swaps persons D and C.

Average speed uses total PATH LENGTH, not displacement, so the turning point must be located first. Here $v = \frac{ds}{dt} = 2t - 6$, which vanishes at $t = 3$ s. Positions are $s(0) = 5\,\text{m}$, $s(3) = -4\,\text{m}$, $s(6) = 5\,\text{m}$, giving 9 m travelled in each leg and 18 m in total. Hence average speed $= 18/6 = 3\ \text{ms}^{-1}$. NCERT Class 11, Motion in a Straight Line.

Using v$^{2}$ = u$^{2}$ + 2gh gives h = $(80^{2} - 20^{2})/(2 \times 10) = 300$ m. Chapter: Motion in a Straight Line.