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NEET PHYSICSMOTION IN A STRAIGHT LINEEasy

Question

Equation of displacement for any particle is s=3t3+7t2+14t+8s = 3t^3 + 7t^2 + 14t + 8 m. Its acceleration at time t=1t = 1 sec is:

A

10 m/s²

B

16 m/s²

C

25 m/s²

D

32 m/s²

Step-by-Step Solution

  1. Velocity (vv): Instantaneous velocity is the rate of change of displacement, v=dsdtv = \frac{ds}{dt} . Given s=3t3+7t2+14t+8s = 3t^3 + 7t^2 + 14t + 8. Differentiating with respect to tt: v=ddt(3t3+7t2+14t+8)=9t2+14t+14v = \frac{d}{dt}(3t^3 + 7t^2 + 14t + 8) = 9t^2 + 14t + 14.
  2. Acceleration (aa): Instantaneous acceleration is the rate of change of velocity, a=dvdta = \frac{dv}{dt} . Differentiating vv with respect to tt: a=ddt(9t2+14t+14)=18t+14a = \frac{d}{dt}(9t^2 + 14t + 14) = 18t + 14.
  3. Calculation at t=1t = 1 sec: Substitute t=1t = 1 into the acceleration equation: a=18(1)+14=32 m/s2a = 18(1) + 14 = 32 \text{ m/s}^2.

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 LINEequationdisplacementparticleacceleration

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