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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

A particle starts from rest, accelerates at 2 m/s22 \text{ m/s}^2 for 10 s10 \text{ s} and then goes for constant speed for 30 s30 \text{ s} and then decelerates at 4 m/s24 \text{ m/s}^2 till it stops. What is the distance travelled by it?

A

750 m

B

800 m

C

700 m

D

850 m

Step-by-Step Solution

The motion is divided into three parts:

  1. Acceleration Phase: Initial velocity u=0u = 0, Acceleration a1=2 m/s2a_1 = 2 \text{ m/s}^2, Time t1=10 st_1 = 10 \text{ s}. Distance S1=ut+12at2=0+12(2)(10)2=100 mS_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(10)^2 = 100 \text{ m} .
  • Final velocity reached v=u+at=0+2(10)=20 m/sv = u + at = 0 + 2(10) = 20 \text{ m/s}.
  1. Constant Speed Phase: Velocity v=20 m/sv = 20 \text{ m/s}, Time t2=30 st_2 = 30 \text{ s}. Distance S2=velocity×time=20×30=600 mS_2 = \text{velocity} \times \text{time} = 20 \times 30 = 600 \text{ m}.

  2. Deceleration Phase: Initial velocity u=20 m/su' = 20 \text{ m/s}, Final velocity v=0v' = 0, Acceleration a2=4 m/s2a_2 = -4 \text{ m/s}^2. Using v2u2=2aSv^2 - u^2 = 2aS: 02(20)2=2(4)S30^2 - (20)^2 = 2(-4)S_3.

  • 400=8S3    S3=50 m-400 = -8S_3 \implies S_3 = 50 \text{ m} .

Total Distance: Stotal=S1+S2+S3=100+600+50=750 mS_{total} = S_1 + S_2 + S_3 = 100 + 600 + 50 = 750 \text{ m}.

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 LINEparticlestartsacceleratesconstantdecelerates

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