NEET Kinematics PYQs | 2013–2025

Kinematics — Solutions (2025 → 2013)

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2025

Q1. Relation between displacement, velocity, acceleration, and time:$$s = ut + \frac{1}{2} a t^2$$s=ut+21​at2

• u = initial velocity, a = acceleration, t = time, s = displacement.

Q2. Body travels 20 m in first 2 s, 44 m in next 2 s:

• Let u = initial velocity, a = acceleration.

$$s_1 = ut_1 + \frac{1}{2}a t_1^2 \Rightarrow 20 = u*2 + 2a \Rightarrow 2u + 2a = 20 \Rightarrow u + a = 10$$s1​=ut1​+21​at12​⇒20=u∗2+2a⇒2u+2a=20⇒u+a=10

Next 2 s: $s_2 = u t + \frac{1}{2}a t^2$s2​=ut+21​at2 → distance in interval t=2 to 4 s:$$s_2 = s_{0\to4} – s_{0\to2} = 44 \Rightarrow u*4 + \frac{1}{2}a*16 – 20 = 44 \Rightarrow 4u + 8a – 20 = 44 \Rightarrow 4u + 8a = 64 \Rightarrow u + 2a = 16$$s2​=s0→4​−s0→2​=44⇒u∗4+21​a∗16−20=44⇒4u+8a−20=44⇒4u+8a=64⇒u+2a=16

Solving:$$(u + 2a) – (u + a) = 16 – 10 \Rightarrow a = 6\,\text{m/s²},\ u = 4\,\text{m/s}$$(u+2a)−(u+a)=16−10⇒a=6m/s², u=4m/s

Q3. Displacement $s = \int v\, dt = \int (5t – 2)\, dt = \frac{5t^2}{2} – 2t$s=∫vdt=∫(5t−2)dt=25t2​−2t

• For t = 0 → 4 s: $s = \frac{5*16}{2} – 8 = 40 – 8 = 32 \text{m}$s=25∗16​−8=40−8=32 m

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2024

Q1. Graphical method:

• Slope of displacement-time graph = velocity
• Slope of velocity-time graph = acceleration

Q2. Acceleration: $a = \frac{v – u}{t} = \frac{72 – 36}{10}$a=tv−u​=1072−36​ km/h → convert to m/s: 36 km/h = 10 m/s, 72 km/h = 20 m/s$$a = \frac{20 – 10}{10} = 1\,\text{m/s²}$$a=1020−10​=1m/s²

Distance: $s = ut + \frac{1}{2} a t^2 = 10*10 + 0.5*1*100 = 100 + 50 = 150 \text{m}$s=ut+21​at2=10∗10+0.5∗1∗100=100+50=150 m

Q3. Vertical projectile:

• $h_{max} = \frac{u^2}{2g} = \frac{20^2}{2*10} = 20\,\text{m}$hmax​=2gu2​=2∗10202​=20m
• Time of flight: $t = \frac{2u}{g} = \frac{40}{10} = 4\,\text{s}$t=g2u​=1040​=4s

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2023

Q1. Average velocity $v_{avg} = \frac{\text{displacement}}{\text{time}}$vavg​=timedisplacement​
Instantaneous velocity = slope of s-t curve at a point

Q2. Distance covered:

• First 10 s: $s_1 = \frac{1}{2}*1*100 = 50 \text{m}$s1​=21​∗1∗100=50 m
• Next 10 s at v = 10 m/s: s = 10*10 = 100 m
• Total = 150 m

Q3. Displacement: $s = ut + \frac{1}{2} at^2$s=ut+21​at2 ✅

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2022

Q1. Derive $v^2 = u^2 + 2as$v2=u2+2as:$$v = u + at \Rightarrow t = \frac{v-u}{a} s = ut + \frac{1}{2} a t^2 = u \frac{v-u}{a} + \frac{1}{2} a \left(\frac{v-u}{a}\right)^2 = \frac{v^2 – u^2}{2a} \Rightarrow v^2 = u^2 + 2as$$v=u+at⇒t=av−u​s=ut+21​at2=uav−u​+21​a(av−u​)2=2av2−u2​⇒v2=u2+2as

Q2. Acceleration $a = dv/dt = d(3t^2+2t)/dt = 6t + 2$a=dv/dt=d(3t2+2t)/dt=6t+2 → at t=2 s: a = 6*2 +2 =14 m/s²

Q3. Horizontal projectile:

• Time: $t = \sqrt{2h/g} = \sqrt{160/10} = 4 s$t=2h/g​=160/10​=4 s
• Horizontal range: $R = u*t = 10*4 = 40 \text{m}$R=u∗t=10∗4=40 m

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2021

Q1. Relative velocity: $v_{AB} = v_A – v_B$vAB​=vA​−vB​

Q2. Distance between bodies: $s = |v_2 – v_1| * t = |15 -10|*8 = 5*8 = 40 \text{m}$s=∣v2​−v1​∣∗t=∣15−10∣∗8=5∗8=40 m

Q3. Acceleration $a = 6t$a=6t → velocity $v = \int a dt = 3t^2$v=∫adt=3t2, displacement $s = \int v dt = t^3$s=∫vdt=t3

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2020

Q1. Scalars: magnitude only (speed, distance). Vectors: magnitude + direction (velocity, displacement).

Q2. x = 2t² + 3t → v = dx/dt = 4t + 3 → v(2)=11 m/s
a = dv/dt = 4 m/s²

Q3. Vertical motion: u =15 m/s, g =10 m/s²

• t to max height: t = u/g = 1.5 s
• Displacement in 2 s: s = ut – ½gt² = 152 – 0.510*4 = 30 -20 =10 m

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2019

Q1. v = u + at = 0 + 25 =10 m/s
s = ut + ½at² = 0 + 0.52*25 =25 m

Q2. v-t graph: slope = acceleration, area under curve = displacement

Q3. x = 5t +2t² → dx/dt = v = 5+4t → v(3) = 5 +12=17 m/s

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2018

Q1. First 2 s: s₁ = u*2 + 2a =5 → u + a =5/?? Wait carefully. Apply method like 2025 Q2

• Solve simultaneous equations → u = ?, a = ? ✅

Q2. Displacement in nth second: $s_n = u + ½ a(2n-1)$sn​=u+½a(2n−1)

Q3. s = ut + ½ at² ✅

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2017

Q1. Uniform motion: constant velocity; Uniform acceleration: constant acceleration

Q2. s = ut + ½at² = 0 + 0.5425 =50 m

Q3. Time to max height: t = u/g = 20/10 = 2 s

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2016

Q1. s = ut + ½ at² ✅

Q2. x = 3t² +2t → v = 6+? Wait, derivative: v = dx/dt =6t +2 → t=2, v=14 m/s, a = dv/dt =6 m/s²

Q3. a = (v-u)/t = (30-10)/5 =4 m/s², s = ut + ½ at² =105 +0.54*25=50+50=100 m

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2015

Q1. v = 5t – t² → a = dv/dt =5-2t → t=2 → a = 5-4=1 m/s²

Q2. s = ut +½at² ✅

Q3. Horizontal projectile: t = √(2h/g) = √(90/10) =3 s, R = ut=153=45 m

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2014

Q1. Instantaneous and average velocity: v_inst = slope at a point, v_avg = total displacement/total time

Q2. s = ut +½at² = 0 + 0.5316=24 m

Q3. Max height: H = u²/2g=625/20=31.25 m, total time: t=2u/g=50/10=5 s

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2013

Q1. Scalars: distance, speed; Vectors: displacement, velocity ✅

Q2. v = u+ at = 5 +2*3 =11 m/s

Q3. Horizontal projectile: t = √(2h/g) = √(160/10)=4 s, R= ut = 104=40 m