Exams › NEET › Physics

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

1. α²t²β²/(αβ)
2. αβt/(α+β)
3. (α+β)t
4. αβt/(α+β)

Correct answer: αβt/(α+β)

Solution

The car's motion consists of two phases: acceleration and deceleration. Using the equations of motion, the time spent accelerating is t₁ = v/α, and the time spent decelerating is t₂ = v/β. Since t₁ + t₂ = t, solving for v gives v = αβt / (α + β).

Related NEET Physics questions

• In the equation of motion, \( \boldsymbol{S}=\boldsymbol{u} \boldsymbol{t}+ \) \( 1 / 2 a t^{2}, \) S stands for
• A body is accelerated by applying a force of 30 N. The change in the momentum of the body after 2 sec is?
• A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is
• From the displacement-time graph shown here, find the velocity of the body as it moves from \( A \) to \( B \)
• The distance of a particle as a function of time is shown below. The graph indicates that
• Whenever an object moves with a constant speed, its distance - time graph is a

⚔️ Practice NEET Physics free + battle 1v1 →