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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

If the velocity of a particle is v=At+Bt2v = At + Bt^2, where AA and BB are constants, then the distance travelled by it between 1 s and 2 s is:

A

3A + 7B

B

3A/2 + 7B/3

C

A/2 + B/3

D

3A/2 + 4B

Step-by-Step Solution

Velocity is defined as the rate of change of position with respect to time (v=dxdtv = \frac{dx}{dt}) . To find the distance travelled between two instants, we integrate the velocity function over the given time interval.

  1. Set up the integral: Distance=∫t1t2v dt=∫12(At+Bt2) dt\text{Distance} = \int_{t_1}^{t_2} v \, dt = \int_{1}^{2} (At + Bt^2) \, dt

  2. Integrate: Using the power rule ∫tndt=tn+1n+1\int t^n dt = \frac{t^{n+1}}{n+1}: x=[At22+Bt33]12x = \left[ \frac{At^2}{2} + \frac{Bt^3}{3} \right]_{1}^{2}

  3. Apply limits (Upper limit 2, Lower limit 1): x=(A(2)22+B(2)33)−(A(1)22+B(1)33)x = \left( \frac{A(2)^2}{2} + \frac{B(2)^3}{3} \right) - \left( \frac{A(1)^2}{2} + \frac{B(1)^3}{3} \right) x=(4A2+8B3)−(A2+B3)x = \left( \frac{4A}{2} + \frac{8B}{3} \right) - \left( \frac{A}{2} + \frac{B}{3} \right) x=2A+8B3−A2−B3x = 2A + \frac{8B}{3} - \frac{A}{2} - \frac{B}{3}

  4. Simplify: x=(2A−A2)+(8B3−B3)x = \left( 2A - \frac{A}{2} \right) + \left( \frac{8B}{3} - \frac{B}{3} \right) x=3A2+7B3x = \frac{3A}{2} + \frac{7B}{3}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A STRAIGHT LINEvelocityparticleconstantsdistancetravelled

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