Motion in a Straight Line NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Motion in a Straight Line question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Motion in a Straight Line previous year questions
23 questions. Tap an option to check your answer.
Q1NEET 2021
A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sₙ be the distance travelled by the block in the interval t = n − 1 to t = n. Then, the ratio Sₙ/Sₙ₊₁ is :
Not quite — the answer is B.
Starting from rest with constant acceleration a, distance in the nth second Sₙ = (a/2)(2n − 1), and Sₙ₊₁ = (a/2)(2n + 1). So Sₙ/Sₙ₊₁ = (2n − 1)/(2n + 1); the reverse ratio is greater than 1, impossible for an accelerating block.
Q2NEET 2020 · Q177
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is : (g = 10 m/s²)
Not quite — the answer is D.
v² = u² + 2gh gives 80² = 20² + 2 × 10 × h, so 6400 − 400 = 20h and h = 300 m. Ignoring the initial downward speed would give 320 m.
Q3NEET 2018 · Q34
A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E⃗. Due to the force qE⃗, its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively
Not quite — the answer is B.
a = 6 m/s². 0–1 s: 3 m forward. 1–2 s: decelerates from 6 to 0, another 3 m forward. 2–3 s: accelerates backward from rest, 3 m back. Net displacement 3 m and distance 9 m in 3 s, so average velocity 1 m/s and average speed 3 m/s.
Q4NEET 2017 · Q151
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be:
Not quite — the answer is C.
With escalator length L her speed is L/t₁ and the escalator's is L/t₂. Moving together the speeds add, so t = L/(L/t₁ + L/t₂) = t₁t₂/(t₁ + t₂), which is shorter than either time. The average (t₁ + t₂)/2 wrongly averages times instead of adding speeds.
Q5NEET 2016 Phase 2 · Q2
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) = at + bt² and x_Q(t) = ft − t². At what time do the cars have the same velocity?
Not quite — the answer is D.
Differentiating, v_P = a + 2bt and v_Q = f − 2t. Setting them equal gives 2t(1 + b) = f − a, so t = (f − a)/(2(1 + b)). Writing a + f in the numerator comes from a sign slip when differentiating −t².
Q6NEET 2016 Phase 1 · Q41
If the velocity of a particle is v = At + Bt², where A and B are constants, then the distance travelled by it between 1s and 2s is:
Not quite — the answer is C.
Distance = ∫₁² (At + Bt²) dt = A(2² − 1²)/2 + B(2³ − 1³)/3 = (3/2)A + (7/3)B (the velocity does not change sign for positive A, B). A/2 + B/3 is the result of integrating from 0 to 1 instead.
Q7AIPMT 2015
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx⁻²ⁿ, 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:
Not quite — the answer is C.
a = v dv/dx. With v = βx⁻²ⁿ, dv/dx = −2nβx⁻²ⁿ⁻¹, so a = βx⁻²ⁿ × (−2nβx⁻²ⁿ⁻¹) = −2nβ²x⁻⁴ⁿ⁻¹. Using only dv/dx gives the −2n−1 power, which is not an acceleration.
Q8NEET 2013
A stone falls freely under gravity. It covers distances h₁, h₂ and h₃ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁, h₂ and h₃ is
Not quite — the answer is B.
Starting from rest, distances covered in successive equal time intervals are in the ratio 1 : 3 : 5 (Galileo's law of odd numbers). So h₂ = 3h₁ and h₃ = 5h₁, i.e. h₁ = h₂/3 = h₃/5. Equal distances would hold only for uniform velocity.
Q9AIPMT 2012 Prelims
The motion of a particle along a straight line is described by equation x = 8 + 12t − t³, where x is in metre and t in second. The retardation of the particle when its velocity becomes zero is
Not quite — the answer is A.
v = dx/dt = 12 − 3t², which is zero at t = 2 s. a = dv/dt = −6t = −12 m s⁻² at t = 2 s, so the retardation is 12 m s⁻². Zero is wrong because velocity being zero does not make acceleration zero.
Q10AIPMT 2011 Mains
A particle covers half of its total distance with speed v₁ and the rest half distance with speed v₂. Its average speed during the complete journey is
Not quite — the answer is D.
Each half takes d/v₁ and d/v₂, so average speed = 2d/(d/v₁ + d/v₂) = 2v₁v₂/(v₁ + v₂), the harmonic mean. The arithmetic mean (v₁ + v₂)/2 would apply only to equal time intervals, not equal distances.
Q11AIPMT 2011 Prelims
A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m s⁻², the velocity with which it hits the ground is
Not quite — the answer is C.
For a stone dropped from rest, v² = 2gh = 2 × 10 × 20 = 400, so v = 20 m/s. Forgetting the factor 2 under the root gives about 14 m/s, and v = gh would give an unphysical 200 m/s.
Q12AIPMT 2010 Prelims · Q2
A particle moves a distance x in time t according to equation x = (t + 5)⁻¹. The acceleration of particle is proportional to
Not quite — the answer is B.
With x = (t + 5)⁻¹, v = −(t + 5)⁻² and a = 2(t + 5)⁻³. Since |v| = (t + 5)⁻², (t + 5)⁻³ = |v|^(3/2), so a ∝ (velocity)^(3/2). In terms of distance a = 2x³, so neither (distance)² nor (distance)⁻² fits.
Q13AIPMT 2010 Prelims · Q14
A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (Take g = 10 m/s²)
Not quite — the answer is B.
At t = 18 s the first ball has fallen ½ × 10 × 18² = 1620 m. The second ball moves for 12 s: 12v + ½ × 10 × 12² = 12v + 720. Equating gives 12v = 900, v = 75 m/s.
Q14AIPMT 2009 Prelims · Q35
A bus is moving with a speed of 10 m s⁻¹ on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus?
Not quite — the answer is D.
The scooter must close 1000 m in 100 s, so its speed relative to the bus is 10 m s⁻¹. Its ground speed = 10 + 10 = 20 m s⁻¹. Answering 10 m s⁻¹ gives only the relative speed.
Q15AIPMT 2009 Prelims · Q40
A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is S₁ and that covered in the first 20 seconds is S₂, then:
Not quite — the answer is B.
From rest under constant force, S = ½at², so S ∝ t². Doubling the time from 10 s to 20 s makes S₂ = 4S₁. S₂ = 3S₁ is the distance in the second 10 s, not the total.
Q16AIPMT 2008 Prelims · Q3
The distance travelled by a particle starting from rest and moving with an acceleration (4/3) m s⁻², in the third second is
Not quite — the answer is D.
Distance in the nth second from rest = (a/2)(2n − 1) = (2/3)(2 × 3 − 1) = 10/3 m. 6 m is the total distance covered in the first three seconds, (1/2)(4/3)(3²), not the distance in the third second alone.
Q17AIPMT 2008 Prelims · Q4
A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 m s⁻¹ to 20 m s⁻¹ while passing through a distance 135 m in t second. The value of t is
Not quite — the answer is A.
For uniform acceleration s = (u + v)t/2, so 135 = (10 + 20)t/2 and t = 9 s. Check: a = (20² − 10²)/(2 × 135) = 10/9 m s⁻², and (20 − 10)/(10/9) = 9 s.
Q18AIPMT 2007 Prelims · Q3
The position x of a particle with respect to time t along x-axis is given by x = 9t² − t³ where x is in metres and t in second. What will be the position of this particle when it achieves maximum speed along the +ve x direction?
Not quite — the answer is A.
v = dx/dt = 18t − 3t² and a = 18 − 6t; speed is maximum when a = 0, i.e. t = 3 s. Then x = 9(9) − 27 = 54 m. 81 m is just the 9t² term without subtracting t³.
Q19AIPMT 2007 Prelims · Q5
A car moves from X to Y with a uniform speed vu and returns to X with a uniform speed vd. The average speed for this round trip is
Not quite — the answer is D.
For equal distances d each way, total time = d/vu + d/vd, so average speed = 2d/(d/vu + d/vd) = 2vu vd/(vu + vd), the harmonic mean. The arithmetic mean (vu + vd)/2 is wrong because more time is spent at the lower speed.
Q20AIPMT 2007 Prelims · Q18
A particle moving along x-axis has acceleration f, at time t, given by f = f₀(1 − t/T), where f₀ and T are constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle's velocity (vx) is
Not quite — the answer is C.
f = 0 at t = T. v = ∫₀ᵀ f₀(1 − t/T) dt = f₀T − f₀T/2 = ½f₀T. f₀T would be the velocity if the acceleration stayed at f₀ throughout.
Q21AIPMT 2006 Prelims · Q50
Two bodies, A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is :
Not quite — the answer is D.
In free fall from rest, h = ½gt², so t = √(2h/g) and does not depend on mass. t_A/t_B = √(16/25) = 4/5. Inverting the ratio or including the masses gives the wrong options.
Q22AIPMT 2005 Prelims · Q25
The displacement x of a particle varies with time t as x = ae^(−αt) + be^(βt), where a, b, α and β are positive constants. The velocity of the particle will :
Not quite — the answer is D.
v = −aαe^(−αt) + bβe^(βt) and dv/dt = aα²e^(−αt) + bβ²e^(βt) > 0 for all t, so the velocity keeps increasing. It clearly depends on α and β, and α = β does not make it zero for all times.
Q23AIPMT 2005 Prelims · Q50
A ball is thrown vertically upward. It has a speed of 10 m/s when it has reached one half of its maximum height. How high does the ball rise ? (Taking g = 10 m/s²)
Not quite — the answer is B.
From H/2 to the top the speed drops from 10 m/s to 0: 10² = 2g(H/2) = gH, so H = 100/10 = 10 m. 5 m would be just the upper half.
How often Motion in a Straight Line came up
Questions from this chapter in each paper on this site.
- NEET 2021: 1
- NEET 2020: 1
- NEET 2018: 1
- NEET 2017: 1
- NEET 2016 Phase 2: 1
- NEET 2016 Phase 1: 1
- AIPMT 2015: 1
- NEET 2013: 1
- AIPMT 2012 Prelims: 1
- AIPMT 2011 Mains: 1
- AIPMT 2011 Prelims: 1
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 2
- AIPMT 2008 Prelims: 2
- AIPMT 2007 Prelims: 3
- AIPMT 2006 Prelims: 1
- AIPMT 2005 Prelims: 2
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