Motion in Straight Line NEET Questions - NEET Previous Year Questions Papers (PYQ)

Motion in a Straight Line NEET PYQ Analysis (2015-2025)

A detailed analysis of Motion in a Straight Line NEET chapter-wise PYQ explains that the weightage is likely 2-3 questions (8–12 marks), consistent with previous years' questions (2015-2025).

High-Probability Topics:

• Graphs: Velocity-time (slope = acceleration, area = displacement).

• Equations of Motion: Applications in free fall, deceleration, or non-uniform acceleration.

• Relative Velocity: Problems involving two objects moving in opposite/same directions.

• Average Speed vs. Average Velocity: Conceptual distinctions.

Difficulty Level:

• 60% Medium (application-based), e.g., combining graphs with equations.

• 30% Easy (direct formula-based), e.g., calculating displacement.

• 10% Hard (twist in relative motion or free fall).

Motion in a Straight Line PYQ for NEET 2026: Important Formulas and Derivations

Aspirants are advised to go through some important formulas and derivations before practising from the motion in a straight line NEET question bank:

Some important formulas and derivations are:

Equations of Motion (constant acceleration):

Final velocity: v=u+at

Displacement:

$s = ut + \tfrac{1}{2}at^2$

$v^2 = u^2 + 2as$

Average speed (unequal time intervals): $\dfrac{\text{Total Distance}}{\text{Total Time}}$

Average speed (equal distances): $\dfrac{2v_1v_2}{v_1 + v_2}$

Relative velocity of A with respect to B: $v_{A/B} = v_A - v_B$

Free fall velocity: $v = u + gt \;\; (g = +9.8 \, \text{m/s}^2 \, \text{if downward is positive})$

Free fall height: $h = ut + \tfrac{1}{2}gt^2$

Graphical relations:

- Slope of $s$–$t$ graph = instantaneous velocity

- Slope of $v$–$t$ graph = acceleration

- Area under $v$–$t$ graph = displacement

Motion in a Straight Line NEET Important Questions

Given below are NEET Physics Motion in a Straight Line previous year questions for analysis and practice. Once aspirants are done with studying the concepts, they can start practising from the previous year NEET Physics questions on motion in a straight line.

Question 1: Particle velocity is given by the relation $v = 2e^t + 3e^{2t}$, the acceleration at $t = 0 \, \text{sec}$ will be:

Options:

(1) $5 \, \text{m/s}^2$

(2) $8 \, \text{m/s}^2$

(3) $15 \, \text{m/s}^2$

(4) $6 \, \text{m/s}^2$

Solution: Given velocity
$v = 2e^t + 3e^{2t}$

Acceleration is
$a = \dfrac{dv}{dt} = \dfrac{d}{dt}\left(2e^t + 3e^{2t}\right)$

So,
$a = 2e^t + 6e^{2t}$

At $t = 0 \, \text{sec}$,
$a = 2e^0 + 6e^{2 \times 0} \, (\because e^0 = 1)$

$a = 2 \times 1 + 6 \times 1$

$a = 8 \, \text{m/s}^2$

Hence, the correct answer is option $(2)$.

Question 2: A particle is dropped from a tower. It is found that it travels $45 \, \text{m}$ in the last second of its journey. Find the height of the tower. (Take $g = 10 \, \text{m/s}^2$)

Options:

(1) $200 \, \text{m}$

(2) $125 \, \text{m}$

(3) $370 \, \text{m}$

(4) $120 \, \text{m}$

Solution:
Let the total time of the journey be $n$ seconds.

Displacement in the $n^{th}$ second is:
$S_n = u + \dfrac{a}{2}(2n - 1)$

Given $S_n = 45 \, \text{m}, \; u = 0, \; a = 10 \, \text{m/s}^2$:
$45 = 0 + \dfrac{10}{2}(2n - 1)$

$45 = 5(2n - 1)$

$45 = 10n - 5$

$50 = 10n \;\; \Rightarrow \;\; n = 5$

So, total time of journey = $5$ seconds.

Height of tower:
$h = ut + \dfrac{1}{2}gt^2$

$h = 0 + \dfrac{1}{2} \times 10 \times 5^2$

$h = 5 \times 25 = 125 \, \text{m}$

Ans: Height of the tower is $125 \, \text{m}$. Hence, the correct answer is option (2).

Question 3: An aeroplane is moving with velocity $v(t) = t + \dfrac{2}{t}$, where $t$ is time. When the aeroplane is at its maximum height, it becomes stable. After some time, it returns to the runway with the same velocity. What will be the acceleration at that particular time?

Options:

(1) $\dfrac{1}{t} - \dfrac{1}{t^{3/2}}$

(2) $\dfrac{4}{t} + \dfrac{1}{2t}$

(3) $\dfrac{1}{t} + \dfrac{1}{t^{3/2}}$

(4) $\dfrac{1}{2t} - \dfrac{1}{t^{3/2}}$

Solution:
Velocity: $v = t + \dfrac{2}{t}$

Acceleration:
$a = \dfrac{dv}{dt} = \dfrac{d}{dt}\left(t + \dfrac{2}{t}\right)$

Rewrite: $v = t^{1/2} + 2t^{-1/2}$

Differentiate term by term:
$a = \dfrac{1}{2}t^{-1/2} + 2 \cdot \left(-\dfrac{1}{2}\right)t^{-3/2}$

$a = \dfrac{1}{2t^{1/2}} - \dfrac{1}{t^{3/2}}$

Ans: Hence, the correct answer is option (4).

Question 4: A boy is moving with acceleration $A \propto \beta t n^2$. Acceleration is proportional to ----- when $n = e^2$.

Options:

(1) $te^2$

(2) $te^4$

(3) $te^3$

(4) $te^0$

Solution:
From the problem,
$\dfrac{dV}{dt} = A \propto \beta t n^2 \;\;(1)$

Substitute $n = e^2$ into (1):
$a = A \propto \beta \cdot t (e^2)^2$

$a \propto t e^4$

Ans: Hence, the correct answer is option (2).

Question 5: A particle moves along a straight line such that its displacement at any time $t$ is given by
$s = t^3 - 6t^2 + 3t + 4 \, \text{ m}$.
The velocity when the acceleration is zero:

Options:

(1) $-9 \, \text{m/s}$

(2) $-10 \, \text{m/s}$

(3) $-6 \, \text{m/s}$

(4) $-4 \, \text{m/s}$

Solution: Given
$s = t^3 - 6t^2 + 3t + 4$

Velocity:
$v = \dfrac{ds}{dt} = \dfrac{d}{dt}(t^3 - 6t^2 + 3t + 4) = 3t^2 - 12t + 3 (1)$

Acceleration:
$a = \dfrac{dv}{dt} = \dfrac{d}{dt}(3t^2 - 12t + 3) = 6t - 12 2)$

But acceleration is zero:
$a = \dfrac{dv}{dt} = 0$

From (2):
$0 = 6t - 12 \;\; \Rightarrow \;\; t = 2 \, \text{sec}$

Now, put $t = 2 \, \text{sec}$ in equation (1):
$v = 3(2)^2 - 12 \times 2 + 3$

$v = 12 - 24 + 3$

$v = -12 + 3 = -9 \, \text{m/s}$

Ans: Hence, the correct answer is option (1).

Common Mistakes while Solving Motion in a Straight Line NEET Chapter-wise PYQ

1. Speed ≠ Velocity: Speed is scalar (no direction); velocity is vector (direction matters).

2. Sign Errors in Free Fall: Using g=+9.8 m/s² even when upward is chosen as positive.

3. Graph Confusion: Mixing up displacement-time and velocity-time graphs.

4. Incorrect Average Speed: Assuming Avg. speed =v¹+v² for unequal distances.

5. Relative Velocity Oversights: Forgetting to subtract velocities vectorially (e.g., $v_{\text{rain, man}} = v_{\text{rain}} - v_{\text{man}}$)

6. Deceleration Misinterpretation: Deceleration is acceleration opposite to velocity (sign depends on coordinate system).

7. Ignoring Real-World Factors: Assuming g=10 m/s2 or neglecting air resistance (NEET uses 9.8 m/s² if not mentioned in the question).