Motion In One Dimension And Motion In A Plane: Important Questions for Revision

Motion in One Dimension and Motion in a Plane can be challenging for NEET aspirants due to the use of equations of motion, vector calculations, graphs, projectile motion, and circular motion. Understanding the right formulas and their applications is essential for solving numerical questions accurately.

For NEET 2026, revise key concepts, practise important topic-wise questions, and focus on common problem-solving methods. PW’s NEET 2026 preparation batches can support your preparation with dedicated guidance, classes, and Biology and Physics revision resources.

Motion in One Dimension and Motion in a Plane: Topic-wise Important Formulas

Practise these important topic-wise questions on Motion in One Dimension and Motion in a Plane to strengthen your concepts, improve problem-solving skills, and prepare for NEET 2026.

Average Speed In Two Stages Of Motion

Average speed is defined as:

Average Speed = Total Distance / Total Time

Consider a train that starts from rest and reaches 80 km/h in time t with uniform acceleration. It then moves at 80 km/h for 3t.

First Stage: Uniform Acceleration

For uniform acceleration:

Distance = Average Velocity × Time

Average velocity during the first stage is:

(0 + 80) / 2 = 40 km/h

Therefore:

Distance in first stage = 40t

Second Stage: Constant Speed

Distance in second stage:

80 × 3t = 240t

Total distance:

40t + 240t = 280t

Total time:

t + 3t = 4t

Hence:

Average speed = 280t / 4t

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Average Speed = 70 km/h

Average speed must be calculated using total distance divided by total time. It cannot be found by simply averaging the two listed speeds because the time spent in each stage is different.

Train Crossing A Tunnel

For a train to completely cross a tunnel:

Time = (Tunnel Length + Train Length) / Train Speed

Suppose train A has length l and speed 108 km/h, while train B has length 4l and speed 72 km/h. Convert the speeds:

• 108 km/h = 30 m/s

• 72 km/h = 20 m/s

Therefore:

Time for A = (L + l) / 30

Time for B = (L + 4l) / 20

If train A takes 35 seconds less than train B:

(Time for B) − (Time for A) = 35 seconds

Using the given relation L = 60l, the equation becomes:

64l / 20 − 61l / 30 = 35

Solving:

l = 30 m

Thus:

L = 60 × 30

Tunnel Length = 1800 m

The main challenge is handling different train lengths, speed conversion, and the time difference together. The problem may appear confusing in an examination because two trains and a time relation are introduced simultaneously.

Average Velocity In Circular Motion

Average velocity depends on displacement, not distance:

Average Velocity = Displacement / Total Time

For a particle moving along a circular arc, the displacement is the chord:

Displacement = 2r sin(angle / 2)

For an angular displacement of 120 degrees:

Displacement = 2r sin 60 degrees = √3r

The distance travelled is the arc length:

Arc Length = r × angle in radians

120 degrees = 2π / 3 radians

Therefore:

Arc length = 2πr / 3

If the constant speed is 5 m/s:

Time = Arc Length / Speed = 2πr / 15

Hence:

Average velocity = √3r / (2πr / 15)

Average Velocity = 15√3 / 2π m/s

Interpreting A v² Versus Position Graph

For constant acceleration:

v² = u² + 2ax

A graph of v² versus x has the form:

v² = mx + c

Therefore:

• Slope = 2a

• Intercept = u²

If the graph has slope:

(60 − 40) / (20 − 10) = 2

Then:

2a = 2

Acceleration = 1 m/s²

Graph-Interpretation Method

1. Identify the variables on the two axes.

2. Find the slope of the graph.

3. Compare the graph with v² = u² + 2ax.

4. Equate the slope with 2a.

5. Use the intercept to find u² if required.

Among the three equations of motion, this is the equation containing v².

Projectile Motion At The Highest Point

For a projectile launched with speed u at angle θ from the horizontal:

Maximum Height = u² sin²θ / 2g

At the highest point:

• Vertical velocity is zero.

• Horizontal velocity remains unchanged.

• Vertical momentum is zero.

• Kinetic energy is generally not zero.

• Gravitational potential energy is maximum.

If the angle is measured from the vertical as φ, then θ = 90 degrees − φ. Therefore:

Maximum Height = u² cos²φ / 2g

Thus, always check whether the angle is measured from the horizontal or vertical before applying the formula.

Complementary-Angle Projectiles

Two projectiles launched with the same speed at complementary angles θ and 90 degrees − θ have the same range.

Their important relations are:

• Same range: R1 = R2 = R

• Different maximum heights

• Different times of flight

• H1H2 = R² / 16

• T1T2 = 2R / g

• H1 / H2 = tan²θ

Relative Velocity In Wind

Relative velocity problems require careful identification of the reference frame.

If a butterfly moves northeast with velocity 4√2 m/s relative to the wind, its components are:

Butterfly velocity relative to wind = 4i + 4j

If the wind moves from north to south at 1 m/s:

Wind velocity = −j

Using:

Velocity Of Butterfly Relative To Ground = Velocity Relative To Wind + Wind Velocity Relative To Ground

Ground velocity:

4i + 4j − j = 4i + 3j

Magnitude:

√(4² + 3²) = 5 m/s

In 3 seconds, displacement magnitude is:

5 × 3 = 15 m

The key point is to distinguish velocity relative to the wind from velocity relative to the ground.

Circular Motion And Acceleration

In uniform circular motion:

• Speed remains constant.

• Tangential acceleration is zero.

• Acceleration is directed toward the centre.

• Centripetal acceleration is:

a = v² / r

For a particle moving with speed 4 m/s in a circle of radius 2 m:

a = 16 / 2 = 8 m/s²

Velocity is always tangent to the circle, while acceleration points radially inward.

For a rotating table, a ball moving outward in a smooth radial groove can be analysed using centrifugal acceleration:

a = ω²x

Since a = v dv/dx, integration gives the radial speed as the ball reaches the edge. This type of problem requires a rotating reference frame and careful use of pseudo-force concepts.

Acceleration From A Displacement Relation

If:

x² = 1 + t²

Differentiate once:

2x dx/dt = 2t

Therefore:

xv = t

Differentiate again:

v² + xa = 1

Since v = t/x:

a = (1 − v²) / x

Using x² − t² = 1:

a = 1 / x³

Therefore:

Acceleration = x⁻³

The product rule is essential while differentiating xv.

Important Questions on Motion in One Dimension and Motion in a Plane

Practise these high-yield questions to strengthen your understanding of core concepts, graph analysis, and relative motion in 1D and 2D kinematics.

Question 1

A train starting from rest first accelerates uniformly up to a speed of 80 km/h in time t. It then moves at a constant speed for time 3t. What is the average speed of the train during the entire journey (in km/h)?

(a) 80 km/h 

(b) 70 km/h 

(c) 30 km/h 

(d) 40 km/h

Question 2

In a car race on a straight road, car A takes time t less than car B to reach the finish line and passes the finishing point with a speed v greater than that of car B. Both cars start from rest and travel with constant accelerations a₁ and a₂, respectively. Then, v is equal to:

(a) [2a₁a₂ / (a₁ + a₂)] t 

(b) 2a₁a₂t 

(c) √(a₁a₂) t 

(d) [(a₁ + a₂)/2] t

Question 3

As shown in the figure, a particle moves with a constant speed of π m/s. Considering its motion from point A to point B, what is the magnitude of its average velocity?

(a) π m/s 

(b) 3 m/s 

(c) 2√3 m/s 

(d) 1.5√3 m/s

Note: The figure is required to determine the displacement and identify the correct option.

Question 4

A particle moves in a circle of radius 50 cm such that, at any instant, the normal and tangential components of its acceleration are equal. If its speed at t = 0 is 4 m/s, the time taken to complete the first revolution is given by:

t = (1/α)[1 − e⁻²π] s

Find the value of α.

Question 5

A ball is thrown vertically upward with an initial velocity such that it reaches a maximum height h. Find the ratio of the times taken to reach height h/3 while moving upward and downward, respectively.

(a) (√2 − 1)/(√2 + 1) 

(b) (√3 − √2)/(√3 + √2) 

(c) (√3 − 1)/(√3 + 1) 

(d) 1/√3

Question 6

A NCC parade moves at a uniform speed of 9 km/h under a mango tree on which a monkey is sitting at a height of 19.6 m. At a particular instant, the monkey drops a mango. At what distance from the tree should a cadet be standing at the time of the drop to receive the mango?

Given: g = 9.8 m/s²

(a) 24.5 m 

(b) 19.8 m 

(c) 10 m 

(d) 5 m

Question 7

The velocity of a particle is given by:

v⃗ = −x î + 2y ĵ − z k̂ m/s

What is the magnitude of its acceleration at the point (1, 2, 4), in m/s²?

(a) 6√6 m/s² 

(b) 9 m/s² 

(c) 3√3 m/s² 

(d) 0 m/s²

Question 8

The x- and y-coordinates of a projectile as functions of time t are given by:

x = 24t

y = 43.6t − 4.9t²

What angle does the projectile's velocity make with the horizontal at t = 2 s?

(a) 60° 

(b) 45° 

(c) 30° 

(d) 75°

Revising Motion in One Dimension and Motion in a Plane requires a clear understanding of equations of motion, graphs, vectors, projectile motion, and circular motion. Practise numerical questions regularly and revise key formulas to strengthen your Physics preparation for NEET 2026. For dedicated guidance and focused revision, explore PW’s NEET 2026 preparation batches. 

Related Resource

Motion in a Straight Line NEET Formulae 

Motion in a Straight Line NEET Notes 

Motion in a Straight Line NEET PYQs with Solutions 

Motion in a Plane: Concepts and Important Topics 

Motion in a Plane NEET Formulae