NDA (II) 2024 GAT Q91: A vehicle starts moving along a straight-line path from

Q: What is the answer to NDA (II) 2024 GAT Q91?

A. Correct — A, 10 s. Phase 1 (from rest, a = 2 m/s² for t s): distance s₁ = ½·2·t² = t², and the speed reached is v = 2t. Phase 2 (a = 5 m/s² for 10 s, starting at speed 2t): distance s₂ = (2t)(10) + ½·5·(10)² = 20t + 250. Total s₁ + s₂ = t² + 20t + 250 = 550, so t² + 20t − 300 = 0, i.e. (t + 30)(t − 10) = 0, giving t = 10 s (the negative root is rejected). (Provisional — pending the official UPSC key.)

NDA (II) 2024 General Ability Test · Question 91 of 141

Question text verified against the official NDA question paper.

Provisional answers. UPSC has not yet released the official NDA 2024 answer key (it is published several months after the exam). The answers here are our best-supported answers — derived from careful subject-solving and cross-checked against independent solved papers and primary sources — and will be reconciled to the official key as soon as UPSC publishes it. They are not the official UPSC answer key.

A vehicle starts moving along a straight-line path from rest. In first t seconds it moves with an acceleration of 2 m/s² and then in next 10 seconds it moves with an acceleration of 5 m/s². The total distance travelled by the vehicle is 550 m. The value of time t is

(a)10 s
(b)13 s
(c)20 s
(d)25 s

Answer

Why

Correct — A, 10 s. Phase 1 (from rest, a = 2 m/s² for t s): distance s₁ = ½·2·t² = t², and the speed reached is v = 2t. Phase 2 (a = 5 m/s² for 10 s, starting at speed 2t): distance s₂ = (2t)(10) + ½·5·(10)² = 20t + 250. Total s₁ + s₂ = t² + 20t + 250 = 550, so t² + 20t − 300 = 0, i.e. (t + 30)(t − 10) = 0, giving t = 10 s (the negative root is rejected).

Why the others are wrong

(b)13 s — Substituting t = 13 into t² + 20t + 250 gives 169 + 260 + 250 = 679 m, not 550 m.
(c)20 s — Substituting t = 20 gives 400 + 400 + 250 = 1050 m, far more than 550 m.
(d)25 s — Substituting t = 25 gives 625 + 500 + 250 = 1375 m; the quadratic has only one positive root, 10 s.

Concept

This is a two-stage uniformly-accelerated motion problem solved with s = ut + ½at². The key link is that the final velocity of the first stage becomes the initial velocity (u = 2t) of the second stage. Adding the two distances and setting the sum equal to the given total yields a quadratic in t.

Break the motion into stages and carry the end-speed forward. Set up s₁ (starts from rest) and s₂ (starts at 2t), add them, and equate to 550 m. The negative root of the quadratic is unphysical, leaving t = 10 s. Substituting the options back in (back-solving) is a fast check under exam pressure.

Key facts

Equations of motion (constant a): v = u + at, s = ut + ½at², v² = u² + 2as.
From rest for t s at 2 m/s²: distance = t², end speed = 2t.
Next 10 s at 5 m/s² starting at 2t: distance = 20t + 250.
t² + 20t − 300 = 0 -> t = 10 s.

Study next

Equations of motion for uniform acceleration (kinematics)Velocity-time graphs and area = displacementMulti-stage motion problems (carry-over of final velocity)Distance vs displacement, speed vs velocitySolving quadratic equations and rejecting unphysical roots

Common traps

Forgetting that Stage 2 starts at speed 2t, not from rest.
Choosing the negative root of the quadratic.

Asked as a numerical using s = ut + ½at² across one or two stages of accelerated motion.

Related PYQs

No directly related past PYQ was found.

Practice

practice — not a real PYQ
A body starts from rest and accelerates uniformly at 2 m/s². The distance covered in the first 5 s is
- (a)10 m
- (b)25 m
- (c)50 m
- (d)100 m
Answer(b) 25 m — s = ½·2·5² = 25 m.
practice — not a real PYQ
A car starting from rest reaches 20 m/s in 10 s with uniform acceleration. Its acceleration is
- (a)1 m/s²
- (b)2 m/s²
- (c)5 m/s²
- (d)10 m/s²
Answer(b) 2 m/s² — a = v/t = 20/10.

(a)10 s
(b)13 s
(c)20 s
(d)25 s