Question
A particle starts from rest and accelerates uniformly at
2 m/s² for 5 seconds, then moves with constant speed for 5 seconds and finally decelerates uniformly to rest in 2 seconds. What is the total displacement?Solution
We are given a three-part motion: Uniform acceleration from rest for 5 seconds, with a = 2 m/s² Uniform motion at constant speed for 5 seconds Uniform deceleration to rest in 2 seconds We are to find the total displacement. Displacement during acceleration Initial velocity u = 0, acceleration a = 2 m/s², time t = 5 s Using: sā = ut + ½at² = 0 + ½(2)(5²) = 25 m Final velocity after this phase: v = u + at = 0 + 2Ć5 = 10 m/s Displacement during constant velocity Now the particle moves at v = 10 m/s for 5 s: sā = v Ć t = 10 Ć 5 = 50 m Displacement during deceleration In this phase, the particle slows from 10 m/s to 0 in 2 s Use: sā = ut + ½at² = 10Ć2 + ½(ā5)(2²) But better: since deceleration is uniform: Use: sā = average speed Ć time = (10 + 0)/2 Ć 2 = 10 m Total displacement: s = sā + sā + sā = 25 + 50 + 10 = 85 m
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