NEET Physics Question - Kinematics

A train started from rest from a station and accelerated at \(2\text{ ms}^{-2}\) for 10 s. Then, it ran at constant speed for 30 s and thereafter it decelerated at \(4\text{ ms}^{-2}\) until it stopped at the next station. The distance between two station is

650 m

700 m

750 m

800 m

Solution:

Max velocity \( v = 20 \text{ ms}^{-1}\). Acceleration distance \(s_1 = \frac{1}{2}(2)(10^2) = 100\text{ m}\). Constant velocity distance \(s_2 = 20 \times 30 = 600\text{ m}\). Deceleration distance \(s_3 = \frac{20^2}{2 \times 4} = 50\text{ m}\).

Total distance = 750 m.

Rankers Physics

Total Distance between Stations

Solution:

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