Practice question
Question
A particle moves at 18m/s and decelerates at 4m/s2 for 2s, then accelerates at 3m/s2 until its speed is 18m/s. What is the total distance covered?
Explanation
Phase 1: v=18−4⋅2=10m/s, x1=18⋅2−12⋅4⋅(2)2=36−8=28m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 65 m as the result, so option A is correct.
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