NEET PhysicsUpdated 19 Oct 2026

Motion in a Straight Line: NEET PYQs and weightage

Motion in a Straight Line carries 2.8 of 100 Physics marks in NEET by past-paper share: 21 of the 758 tagged NEET Physics PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 20 of 30 Physics chapters.

Take the 25-question Motion in a Straight Line test Formula sheet ↓All NEET Physics chapters

2.8/100 Weight in Physics

Exam importance (3/5, by past-paper share)

21 NEET PYQs

369 Practice questions

348 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Physics PYQs in the bank (21 of 758), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Motion in a Straight Line: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

First equation
$v = v_{0} + a t$
Use it when: time is given or asked and the distance is not involved (constant $a$ only)
Second equation
$x - x_{0} = v_{0} t + \frac{1}{2} a t^{2}$
Use it when: you need the displacement after time $t$ and the final velocity is not needed
Third equation
$v^{2} = v_{0}^{2} + 2 a (x - x_{0})$
Use it when: time is neither given nor asked
Average-velocity form
$x - x_{0} = \frac{1}{2} (v_{0} + v) t$
Use it when: both end velocities and the time are known and $a$ is not (valid only for constant $a$ )
Average velocity
$v_{avg} = \frac{Δ x}{Δ t} = \frac{x _{2} - x _{1}}{t _{2} - t _{1}}$
Use it when: the question asks for the average over an interval. Average speed is total path $\div$ time, and it is always $\geq ∣ v_{avg} ∣$
Instantaneous velocity
$v = \frac{d x}{d t}$
Use it when: $x$ is given as a function of $t$ . It equals the slope of the tangent on the $x$ - $t$ graph
Instantaneous acceleration
$a = \frac{d v}{d t}$
Use it when: $v$ is given as a function of $t$ . It equals the slope of the tangent on the $v$ - $t$ graph
Acceleration in terms of x
$a = v \frac{d v}{d x}$
Use it when: $v$ is given as a function of $x$ (differentiate to get $a$ ), or $a$ is given and $v$ is wanted at a position $x$ (integrate $v d v = a d x$ )
Varying acceleration
$v = v_{0} + \int_{0}^{t} a d t$
$x = x_{0} + \int_{0}^{t} v d t$
Use it when: $a$ changes with time. The constant- $a$ equations then fail
Area under the v-t graph
$Area under the v - t graph = displacement$
Use it when: a graph is given. Area below the time axis counts as negative displacement
Free fall from rest
$v = - g t$
$y = - \frac{1}{2} g t^{2}$
$v^{2} = - 2 g y (upward taken as +, a = - g)$
Use it when: a body is dropped. In magnitudes this is speed $g t$ after falling $\frac{1}{2} g t^{2}$
Thrown straight up with speed v₀
$time to the top = \frac{v _{0}}{g}$
$maximum height = \frac{v _{0}^{2}}{2 g}$
Use it when: a body is thrown up. At the top $v = 0$ but $a$ is still $g$ downward
Galileo's odd-number law
$distances in successive equal intervals = 1 : 3 : 5 : 7 : \dots$
Use it when: a body falls from rest and the question asks for the distance in the $n$ th second or $n$ th interval
Stopping distance
$d_{s} = \frac{v _{0}^{2}}{2∣ a ∣}$
$so$
$d_{s} \propto v_{0}^{2}$
Use it when: brakes are applied. Doubling the speed gives $4 \times$ the stopping distance. Convert $km h^{- 1}$ to $m s^{- 1}$ (multiply by $\frac{5}{18}$ ) first

Trap

Students plug numbers in without signs. Pick the positive direction first, then give every quantity its sign. With upward taken as positive, a = −g on the way up, at the top and on the way down. The sign of a alone does not tell you whether the body is speeding up or slowing down.

How does NEET ask Motion in a Straight Line?

8 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1PhysicsNEET UG PYQInstantaneous Speed and Velocity

Try right here · No sign-in

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by

X_{P} (t) = a t + b t^{2}

and

X_{Q} (t) = f t - t^{2}

. At what time do the cars have the same speed?

Answer key (C) & worked solution

Correct answer: Option C

X_{P} (t) = a t + b t^{2}; X_{q} (t) = f t - t^{2}

V_{P} = a + 2 b t; V_{Q} = f - 2 t

V_{P} = V_{Q}

a + 2 b t = f - 2 t

\Rightarrow t = \frac{f - a}{2 ( 1 + b )}

Q2PhysicsNEET UG PYQKinematic Equations for Uniformly Accelerated Motion

A stone falls freely under gravity. It covers distances h

_{1}

, h

_{2}

and h

_{3}

in the first 5 second, the next 5 second and the next 5 second respectively. The relation between h

_{1}

, h

_{2}

and h

_{3}

_{1}

=2h

_{2}

=3h

_{3}

h_{1} = \frac{h _{2}}{3} = \frac{h _{3}}{5}

h_{3} = 3 h_{1} an d h_{3} = 3 h_{2}

h_{1} = h_{2} = h_{3}

Answer key (B) & worked solution

Correct answer: Option B

Distance covered in ﬁrst 5 sec

h_{1} = 0 + \frac{1}{2} g (5)^{2}

h_{1} = \frac{25 g}{2}

Distance covered in first 10 sec

s_{2} = 0 + \frac{1}{2} g (10)^{2} = \frac{100 g}{2}

........(1) So distance covered in second 5 sec

h_{2} = s_{2} - h_{1} = \frac{100 g}{2} - \frac{25 g}{2} = \frac{75 g}{2}

…… (2) distance covered in first 15 sec

s_{3} = 0 + \frac{1}{2} g (15)^{2} = \frac{225 g}{2}

So distance covered in last 5 sec

h_{3} = s_{3} - s_{2} = \frac{225 g}{2} - \frac{100 g}{2} = \frac{125 g}{2}

………..(3) Using (1), (2) ,(3) equation

\frac{h _{1}}{\frac{25 g}{2}} = \frac{h _{2}}{\frac{75 g}{2}} = \frac{h _{3}}{\frac{125 g}{2}}

h_{1} = \frac{h _{2}}{3} = \frac{h _{3}}{5}

Q3PhysicsNEET UG PYQKinematic Equations for Uniformly Accelerated Motion

A bullet from a gun is fired on a rectangular wooden block with velocity

u

. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes

\frac{u}{3}

. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is:

A24 cm

B28 cm

C30 cm

D27 cm

Answer key (D) & worked solution

Correct answer: Option D

v^{2} = u^{2} + 2 a s

(\frac{u}{3})^{2} = u^{2} - 2 a x

2 a x = u^{2} - \frac{u ^{2}}{9}

2 a x = \frac{8 u ^{2}}{9} .... (1)

Similarly from starting $\begin{gathered} {v^2} = {u^2} + 2ax \\ 0 = {u^2} - 2a{x_2} \\ 2a{x_2} = {u^2}\,\,\,\,\,\,\,...\left( 2 \right) \\ \end{gathered} $ By (1)/(2)

\frac{x}{x _{2}} = \frac{8}{9}

\frac{24}{x _{2}} = \frac{8}{9}

x_{2} = 27 c m

Q4PhysicsNEET UG PYQPosition, Path length and Displacement

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

A6.25%

B20%

C26.8%

D12.5%

Answer key (C) & worked solution

Correct answer: Option C

Efficiency of ideal heat engine,

η = (1 - \frac{T _{2}}{T _{1}})

_{2}

: Sink temperature T

_{1}

: Source temperature

% η = (1 - \frac{T _{2}}{T _{1}}) \times 100

= (1 - \frac{273}{373}) \times 100

= (\frac{100}{373}) \times 100 = 26.8%

Q5PhysicsNEET UG PYQRelative Motion in One Dimension

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t

_{1}

. On other day, if she remains stationary on the moving escalator, then the escalator takes her up in time t

_{2}

. The time taken by her to walk up on the moving escalator will be

\frac{t _{1} t _{2}}{t _{2} + t _{1}}

t_{1} - t_{2}

\frac{t _{1} + t _{2}}{2}

\frac{t _{1} t _{2}}{t _{2} - t _{1}}

Answer key (A) & worked solution

Correct answer: Option A

Velocity of girl w.r.t. elevator

= \frac{d}{t _{1}} = v_{g e}

Velocity of elevator w.r.t. ground

v_{e G} = \frac{d}{t _{2}}

then velocity of girl w.r.t. ground

v_{g G} = v_{g e} + v_{e G}

i.e.,

v_{g G} = v_{g e} + v_{e G}

\frac{d}{t} = \frac{d}{t _{1}} + \frac{d}{t _{2}}

\frac{1}{t} = \frac{1}{t _{1}} + \frac{1}{t _{2}}

t = \frac{t _{1} t _{2}}{( t _{1} + t _{2} )}

Q6PhysicsNEET UG PYQAcceleration

In some appropriate units, time (t) and position

(x)

relation of a moving particle is given by

t = x^{2} + x

. The acceleration of the particle is

+ \frac{2}{( x + 1 ) ^{3}}

+ \frac{2}{2 x + 1}

- \frac{2}{( x + 2 ) ^{3}}

- \frac{2}{( 2 x + 1 ) ^{3}}

Answer key (D) & worked solution

Correct answer: Option D

t = x^{2} + x

Derivative w.r.t

x

\frac{d t}{d x} = 2 x + 1

∴ \frac{d x}{d t} = \frac{1}{2 x + 1}

∴ v = \frac{1}{2 x + 1}

Derivative w.r.t. time t

∴ a = \frac{- 1}{( 2 x + 1 ) ^{2}} \frac{2 d x}{d t}

= \frac{- 1}{( 2 x + 1 ) ^{2}} 2 v

= \frac{- 1}{( 2 x + 1 ) ^{2}} 2 \frac{1}{( 2 x + 1 )}

= \frac{- 2}{( 2 x + 1 ) ^{3}}

Q7PhysicsNEET UG PYQAcceleration

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to ν(x) = βx - 2

^{n}

, Where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by:

- 2 n β^{2} x^{- 4 n + 1}

- 2 n β^{2} x^{- 2 n - 1}

- 2 n β^{2} x^{- 4 n - 1}

- 2 n β^{2} x^{- 2 n + 1}

Answer key (C) & worked solution

Correct answer: Option C

ν(x) = βx - 2

^{n}

α = \frac{d v}{d t} = \frac{d v}{d x} . \frac{d x}{d t} = \frac{d v}{d x .} v

\frac{d v}{d x} = - 2 n β x^{2 n - 1} . α = - 2 n β x^{- 2 n - 1} . β x^{- 2 n}

= - 2 n β^{2} x^{- 4 n - 1}

Q8PhysicsNEET UG PYQKinematic Equations for Uniformly Accelerated Motion

If the velocity of a particle is v = At + Bt

^{2}

, where A and B are constants, then the distance travelled by it between 1s and 2s is

\frac{3}{2}

A + 4 B

3 A + 7 B

\frac{3}{2}

A

\frac{7}{3}

B

\frac{A}{2} + \frac{B}{3}

Answer key (C) & worked solution

Correct answer: Option C

v = A t + B t^{2}

\Rightarrow \frac{d x}{d t} = A t + B t^{2}

\Rightarrow 0 \int x d x = 1 \int 2 (A t + B t^{2}) d t

\Rightarrow x = \frac{A}{2} (2^{2} - 1^{2}) + \frac{B}{3} (2^{3} - 1^{3})

= \frac{3 A}{2} + \frac{7 B}{3}

Which topics in Motion in a Straight Line carry the most questions?

All 7 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
Kinematic Equations for Uniformly Accelerated Motion	10	140
Average Velocity and Average Speed	4	74
Acceleration	3	65
Relative Motion in One Dimension	2	39
Instantaneous Speed and Velocity	1	30
Position, Path length and Displacement	1	20
Miscellaneous Examples	0	1

Why does Motion in a Straight Line show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Motion in a Straight Line, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

How ViJEEta scores you

What happens when you test Motion in a Straight Line?

Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.

Start the Motion in a Straight Line test

Chapters either side of Motion in a Straight Line by weight

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Motion in a Straight Line: NEET PYQs and weightage

Motion in a Straight Line: formula sheet for NEET

First equation

Second equation

Third equation

Average-velocity form

Average velocity

Instantaneous velocity

Instantaneous acceleration

Acceleration in terms of x

Varying acceleration

Area under the v-t graph

Free fall from rest

Thrown straight up with speed v₀

Galileo's odd-number law

Stopping distance

How does NEET ask Motion in a Straight Line?

Which topics in Motion in a Straight Line carry the most questions?

Why does Motion in a Straight Line show "–" until you answer 3 questions?

What happens when you test Motion in a Straight Line?

Chapters either side of Motion in a Straight Line by weight