Explanation Given: The train acceleratesat 1.1 m/s 2 until it reaches 95 km/h . Then, it moves with uniform velocity then retards at 2 m/s 2 until it comes to rest in the stop. The train stops for 22 s at each station. Formula Used: Equations of motion: v = u + a t s = u t + 1 2 a t 2 v 2 = u 2 + 2 a s Where, v is the final velocity, u is the initial velocity, a is the acceleration, s is the displacement and t is the time. Calculation: Convert the velocity from km/h to m/s. The maximum velocity v is 95 km hr 95 × 1000 3600 = 26.4 m/s (a) Distance between two trips is 15 km. Each station is 3 km apart. Let t 1 be the timefor reaching the maximum velocity, t 2 is a timefor moving with uniform velocity and t 3 is a timefor retarding between the train stops. v = u + a t 1 26.4 = 0 + 1.1 t 1 t 1 = 26.4 1.1 t 1 = 24 s …… (1) The total distance is 15000 m. The distance covered during accelerated motion: s = u t + 1 2 a t 2 d 1 = 0 + 1 2 a 1 t 1 2 d 1 = 1 2 × 1.1 × ( 24 ) 2 = 316.8 m Calculate the distance travelled during the deceleration to stop at the next station. Final velocity would be zero and initial velocity would be 26.4 m/s . v 2 = u 2 + 2 a s 0 = 26.4 2 − 2 × 2 × d 3 d 3 = 174.24 m Time required to travel 174

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ISBN: 2818440037979

Author: GIANCOLI

Publisher: PEARSON

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Chapter 2, Problem 72GP

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Time taken by a train to make a 15km trip in two situations.

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Chapter 2, Problem 71GP

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