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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

A car accelerates from rest at a constant rate α\alpha for some time, after which it decelerates at a constant rate β\beta and comes to rest. If the total time elapsed is tt, then the maximum velocity acquired by the car is:

A

(α2+β2αβ)t\left( \frac{\alpha^2 + \beta^2}{\alpha\beta} \right) t

B

(α2β2αβ)t\left( \frac{\alpha^2 - \beta^2}{\alpha\beta} \right) t

C

(α+β)tαβ\frac{(\alpha + \beta) t}{\alpha\beta}

D

αβtα+β\frac{\alpha\beta t}{\alpha + \beta}

Step-by-Step Solution

  1. Define Variables: Let the maximum velocity acquired be vmaxv_{max}. Let the time taken to accelerate be t1t_1 and the time taken to decelerate be t2t_2. The total time is t=t1+t2t = t_1 + t_2.
  2. Analyze Acceleration Phase: The car starts from rest (u=0u=0) and accelerates at α\alpha. Using the first equation of motion (v=u+atv = u + at) : vmax=0+αt1    t1=vmaxαv_{max} = 0 + \alpha t_1 \implies t_1 = \frac{v_{max}}{\alpha}
  3. Analyze Deceleration Phase: The car starts from vmaxv_{max} and decelerates at β\beta to rest (v=0v=0). 0=vmaxβt2    vmax=βt2    t2=vmaxβ0 = v_{max} - \beta t_2 \implies v_{max} = \beta t_2 \implies t_2 = \frac{v_{max}}{\beta}
  4. Combine and Solve: Substitute t1t_1 and t2t_2 into the total time equation: t=t1+t2=vmaxα+vmaxβt = t_1 + t_2 = \frac{v_{max}}{\alpha} + \frac{v_{max}}{\beta} t=vmax(1α+1β)=vmax(α+βαβ)t = v_{max} \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) = v_{max} \left( \frac{\alpha + \beta}{\alpha\beta} \right) vmax=αβtα+βv_{max} = \frac{\alpha\beta t}{\alpha + \beta}

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 LINEacceleratesconstantdeceleratesconstantelapsed

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