Motion in a Straight Line — Previous Year Questions | eduPhysics
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Class XI · Chapter 02

Motion in a Straight Line — Exam Practice

Practice questions in CBSE, NEET, and JEE Main style covering velocity, acceleration, kinematic equations, and motion graphs.

📝 10 questions 🎯 CBSE · NEET · JEE Main
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.

1CBSE Board Exam Style

CBSE · 1 Mark
Q1. Can a body have zero velocity and yet be accelerating at that same instant? Give an example.
Answer Yes. A ball thrown vertically upward has exactly zero velocity at its highest point, but the acceleration due to gravity (g, directed downward) continues acting on it throughout — including at that instant.
CBSE · 3 Marks
Q2. Using a velocity-time graph, derive the kinematic equation v² = u² + 2as for motion with constant acceleration.
Answer On a v-t graph, the area under the line between t=0 and t gives displacement s — a trapezium with parallel sides u and v, height t: s = ½(u+v)t.

From the first equation, v = u+at → t = (v−u)/a. Substituting: s = ½(u+v)(v−u)/a = (v²−u²)/(2a).

Rearranging: v² = u² + 2as.
CBSE · 5 Marks
Q3. (a) Using the velocity-time graph method, derive all three kinematic equations for motion with constant acceleration. (b) A car moving at 20 m/s is brought to rest by applying brakes over 5 s. Find its retardation and the distance travelled before stopping.
Answer (a) From the v-t graph's slope: a = (v−u)/t → v = u + at. From the area under the graph (as in Q2): s = ut + ½at² (splitting the trapezium into a rectangle and triangle). Eliminating t between these two gives v² = u² + 2as.

(b) u = 20 m/s, v = 0, t = 5 s. a = (v−u)/t = (0−20)/5 = −4 m/s² → retardation = 4 m/s².
s = ut + ½at² = (20×5) + ½(−4)(25) = 100 − 50 = 50 m.

2NEET Style (Single-Correct MCQ)

NEET Style
Q4. A body moving with uniform acceleration has velocity 10 m/s at t=2s and 20 m/s at t=4s. Its acceleration is:
2.5 m/s²
✓ 5 m/s²
10 m/s²
15 m/s²
Solution a = Δv/Δt = (20−10)/(4−2) = 10/2 = 5 m/s².
NEET Style
Q5. The area under a velocity-time graph gives:
Acceleration
✓ Displacement
Jerk
Average speed only
Solution Since v = ds/dt, integrating (finding area under the curve) recovers displacement over that time interval.
NEET Style
Q6. If a particle's position-time graph is a straight line (not horizontal), its velocity is:
Zero
✓ Constant
Continuously increasing
Continuously decreasing
Solution The slope of a position-time graph gives velocity — a straight line has constant slope, hence constant velocity.
NEET Style
Q7. A particle covers equal distances in equal time intervals while moving in a straight line. Its acceleration is:
Increasing
✓ Zero
Constant but nonzero
Undefined
Solution Equal distances in equal times means constant (uniform) velocity — and constant velocity means exactly zero acceleration.

3JEE Main Style

JEE Main Style · Numerical
Q8. A ball is dropped from a height of 45 m. Find the time it takes to reach the ground, and its velocity just before impact. (g = 10 m/s²)
Solution h = ½gt² → 45 = ½(10)t² → t² = 9 → t = 3 s.

v = gt = 10 × 3 = 30 m/s.
JEE Main Style · MCQ
Q9. For a particle undergoing uniformly accelerated motion, the velocity-time graph is:
A horizontal line
✓ A straight, sloped line
A parabola
A vertical line
Solution v = u + at is linear in t — a straight line with slope a (nonzero, since acceleration is nonzero and constant).
JEE Main Style · Numerical
Q10. A particle starts with initial velocity 5 m/s and moves with constant acceleration 2 m/s². Find the distance it travels during the 3rd second.
Solution sn = u + (a/2)(2n−1) = 5 + (2/2)(2×3−1) = 5 + (1)(5) = 10 m.
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Motion in a Plane

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