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Physics · chapter PYQs

Motion in a Straight Line, on the exam screen.

4 questions from NEET 2026, Re-NEET 2026, NEET 2020, 4 minutes, marked +4 / −1. See them with solutions.

Motion in a Straight Line · PYQs

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  1. Question 1 · NEET 2026 · Q15 · +4 / −1

    The following plots show variation of velocity (vv) with time (tt) of a ball thrown vertically upward, and falling back. Which of the following plots is/are correct?

    figure
    Answer and explanation

    Answer: (1)

    Throughout the flight the only acceleration is g directed downward, so dv/dtdv/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.

    figure

  2. Question 2 · NEET 2026 · Q26 · +4 / −1

    When a ruler falls vertically, 5 different persons catch it with different reaction times (g=9.8g = 9.8 m s−2^{-2}). Person A has a reaction time of 0.20 s, Person B has 0.22 s, Person C has 0.18 s, Person D has 0.19 s, and Person E has 0.21 s. What is the correct order of the distance travelled by the ruler for each person?

    Answer and explanation

    Answer: (3)

    The ruler falls freely, so distance s=12gt2s = \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.

  3. Question 3 · Re-NEET 2026 · Q6 · +4 / −1

    Consider a particle moving along a straight line, whose position as a function of time is given by s(t)=αt2−βt+γs(t) = \alpha t^2 - \beta t + \gamma, where α=1 ms−2\alpha = 1\ \text{ms}^{-2}, β=6 ms−1\beta = 6\ \text{ms}^{-1} and γ=5 m\gamma = 5\ \text{m}. The average speed of the particle, in ms−1\text{ms}^{-1} from t=0t = 0 to t=6t = 6 s is:

    Answer and explanation

    Answer: (3)

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

  4. Question 4 · NEET 2020 · Q149 · +4 / −1

    A ball is thrown vertically downward with a velocity of 20 m s−1^{-1} from the top of a tower. It hits the ground after some time with a velocity of 80 m s−1^{-1}. The height of the tower is (g = 10 m s−2^{-2})

    Answer and explanation

    Answer: (3)

    Using v2^{2} = u2^{2} + 2gh gives h = (802−202)/(2×10)=300(80^{2} - 20^{2})/(2 \times 10) = 300 m. Chapter: Motion in a Straight Line.