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

Motion in a Straight Line NEET PYQ analysis: Comprising of more than 20 lakh candidates competing for restricted seats in medical institutions, the National Eligibility Cum Entrance Test (NEET) is India's main medical entrance examination. There are 45 questions (180 marks) for the NEET physics section.

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1. Motion in a Straight Line NEET PYQ Analysis (2024-2015)
2. Important Formulas and Derivations
3. Motion in Straight NEET Previous Year Questions
4. Common Mistakes and Misconceptions

Among the several chapters, Motion on a Straight Line is a fundamental chapter in Kinematics, necessary for understanding motion-related problems. Since it lays the foundation for higher topics like projectile motion and circular motion, the chapter Motion in a Straight Line is important in the Kinematics unit in NEET Physics. With an annual weightage of 2-3 questions, learning this chapter is essential for achieving 8–12 marks, considerably affecting total results. For NEET aspirants, a strong grasp of this chapter is non-negotiable—it bridges basic mechanics and higher-order concepts, making it a strategic stepping stone in exam preparation.

Motion in a Straight Line NEET PYQ Analysis (2024-2015)

Weightage: Likely 2-3 questions (8–12 marks), consistent with previous years' questions (2024-2015).

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).

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Important Formulas and Derivations

Some important formulas and derivations are:

Equations of Motion (constant acceleration):

• Final velocity: v=u+at

• Displacement: s=ut+(1/2)at2

• Velocity-displacement relation: v2=u2+2as

• Derivation: These are derived by integrating acceleration or using velocity-time graphs.

Average Speed: For unequal time intervals: Total Distance / Total Time.

• For equal distances: (2v1v2)/(v1+v2).

Relative Velocity:

• Velocity of A relative to B: vA−vB (vector subtraction).

Free Fall:

• Velocity: v=u+gt (use g=+9.8 m/s2 if downward is positive).

• Height: h=ut+(1/2)gt2

Graphical Relations:

• Slope of displacement-time graph = Instantaneous velocity.

• Slope of velocity-time graph = Acceleration.

• Area under velocity-time graph = Displacement.

Motion in Straight NEET Previous Year Questions

Question 1: A man is standing $40m$ behind the bus. The bus starts with $1\mathrm{m}/{\sec }^2$ constant acceleration, and at the same instant, the man starts moving at constant speed $9\mathrm{m}/\mathrm{s}$. The time taken by a man to catch the bus?

1. 8 sec

2. 6 sec

3. 8 sec or 10 sec

4. None

Solution:

let after time ' t ' man will catch the bus.
for bus -

$$\begin{array}{rl}& x=x_0+ut+\frac{1}{2}a{t}^2,x=40+0(t)+\frac{1}{2}(1)t^2\\ & x=40+\frac{t^2}{2}\text{...(1)}\\ & \text{For man: }x=9t\text{...(2)}\end{array}$$

From equations (1) and (2),

$$\begin{array}{rl}& 40+\frac{t^2}{2}=9t\\ & 80+t^2=18t\\ & t^2-18t+80=0\end{array}$$

Now,

$$\begin{array}{rl}& t^2-(10+8)t+80=0\\ & t^2-10t-8t+80=0\\ & t(t-10)-8(t-10)=0\\ & (t-10)(t-8)=0\\ & t-10=0,t-8=0\\ & t=10\text{ sec},t=8\text{ sec}\end{array}$$

Hence, the answer is option (3).

Question 2: A particle is dropped from a tower. It's found that it travels 45 m in the last second of its journey. Find out the height of the tower.

$($ take $g=10\mathrm{m}/{\mathrm{s}}^2)$

1. 200 m

2. 125 m

3. 370 m

4. 120 m

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

$$\begin{array}{rl}{S}_n=u+\frac{a}{2}(2n-1)\Rightarrow 45& =0+\frac{10}{2}(2n-1)\\ 45& =5(2n-1)\\ 45& =10n-5\\ 50& =10n\\ n& =5\end{array}$$

Height of the tower

$$\begin{array}{rl}& h=ut+\frac{1}{2}g{t}^2\\ & h=0+\frac{1}{2}\times 10\times 25\\ & h=5\times 25\\ & h=125\mathrm{m}\end{array}$$
Hence, the answer is option (2).

Question 3: An aeroplane is moving with velocity , where t is time. When the aeroplane is at its maximum height, the aeroplane will become stable. After some time, return to the runway with the same velocity. What will be the acceleration at that particular time?

Solution: Velocity of the aeroplane

Acceleration of the aeroplane:

Hence, the answer is option (4).

Question 4: A boy moving with acceleration , acceleration is proportional to _______ when becomes :

Solution: Acceleration is given by the statement -

from question, n becomes , put in (1) -

Hence, the answer is option (2).

Question 5: A Particle is released from rest at point O, as shown in the figure. The acceleration of the particle at point B will be ____

1. $1024\alpha$

2. $1024{\alpha }^2$

3. $10240\alpha$

4. $5012\alpha$

Solution:

Velocity at point $0=0\mathrm{m}/\mathrm{s}$.
Velocity at point $\mathrm{A}=\alpha {\mathrm{t}}^4$
Velocity at point $\mathrm{B}=\alpha {\mathrm{t}}^{10}$
Now,
Acceleration at point $(B)=\frac{d}{dt}$ (velocity at point $B$ )

$\begin{array}{rl}& =\frac{d}{dt}\alpha t^{10}\\ & =10\alpha t^9\end{array}$
Acceleration at point $(B)=10\alpha t^9$
but at point B t $=2\sec$, acceleration will become-

$\begin{array}{rl}\text{ Acceleration (a) }& =10\alpha (2{)}^9\\ & =10\alpha \times 1024\\ & =10240\alpha \end{array}$

Hence, the answer is option (3).

Question 6: A particle moving with acceleration , the velocity between to will be

Solution:

To find velocity, we have to integrate it -

Hence, the answer is option (3).

Question 7: Particle velocity is given by the relation , the Acceleration at will be :

Solution: Given:

Acceleration at will be -

Hence, the answer is option (2).

Question 8: A particle moves along a straight line such that its displacement at any time t is given by

The velocity when the acceleration is zero

Solution: Given

But acceleration is zero-

Now,

put in equation - (1) we get velocity

Hence, the answer is option (1).

Question 9: A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backwards, and so on. Each step is 1 m long and requires 1 s. The time when he will fall into the pit 13 m away from him is:-

1. 13 s

2. 37 s

3. 40 s

4. 42 s

Solution: The time taken to move net 2 steps ( 5 steps forward and 3 steps backward) is 8 sec, and so far, 8 steps he takes 32 s. In the last 5 steps, he will take 5 seconds and fall into the pit.
Therefore, the Total time taken is 32 sec + 5 sec = 37 sec

Hence, the answer is option (2).

Common Mistakes and Misconceptions

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)/2 for unequal distances.

5. Relative Velocity Oversights: Forgetting to subtract velocities vectorially (e.g., vrain, man=vrain−vman ).

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).