1. 3/2A + 4B 2. 3A + 7B 3. 3/2A + 7/3B 4. A/2 + B/3

Answer: 3. 3/2A + 7/3B Key idea: distance is the integral of speed over time (here v≥0 ). s=∫₁²(At+Bt²)dt=A4-1/2+B8-1/3=3/2A+7/3B .

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 :

1. \(\frac{3}{2}A + 4B\)

2. \(3A + 7B\)

3. \(\frac{3}{2}A + \frac{7}{3}B\)

4. \(\frac{A}{2} + \frac{B}{3}\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

Best answer

Answer: 3. \(\frac{3}{2}A + \frac{7}{3}B\)

Key idea: distance is the integral of speed over time (here \(v\ge0\)).

\(s=\int_1^2(At+Bt^2)dt=A\dfrac{4-1}{2}+B\dfrac{8-1}{3}=\dfrac32A+\dfrac73B\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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