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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

A car starts from rest and moves with uniform acceleration 'aa' on a straight road from time t=0t = 0 to t=Tt = T. After that, a constant deceleration brings it to rest. In this process the average speed of the car is:

A

aT4\frac{aT}{4}

B

3aT2\frac{3aT}{2}

C

aT2\frac{aT}{2}

D

aTaT

Step-by-Step Solution

  1. Analyze the Motion: The motion consists of two phases: accelerating from rest to a maximum velocity (vmaxv_{max}) and then decelerating from vmaxv_{max} to rest.
  2. Maximum Velocity: During the acceleration phase (from t=0t=0 to t=Tt=T), the initial velocity u=0u=0 and acceleration is aa. Using the equation v=u+atv = u + at, the maximum velocity reached is vmax=0+aT=aTv_{max} = 0 + aT = aT .
  3. Velocity-Time Graph Method: The velocity-time graph for this motion is a triangle starting from the origin, peaking at vmaxv_{max} at time TT, and dropping to zero at total time ttotalt_{total}.
  • The total distance covered (SS) is the area under the v-t graph: Area = 12×base×height=12×ttotal×vmax\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t_{total} \times v_{max} .
  1. Calculate Average Speed: Average speed is defined as the total distance divided by the total time.
  • vavg=Sttotal=12×ttotal×vmaxttotal=vmax2v_{avg} = \frac{S}{t_{total}} = \frac{\frac{1}{2} \times t_{total} \times v_{max}}{t_{total}} = \frac{v_{max}}{2}.
  1. Substitute Value: Substituting vmax=aTv_{max} = aT, we get:
  • vavg=aT2v_{avg} = \frac{aT}{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 LINEstartsuniformaccelerationstraightconstant

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