Practice question
Question
A cyclist accelerates from 6 m/s at 1.5 m/s² for 5 s, then decelerates at 2 m/s² until its speed is 8 m/s . What is the total distance covered?
Explanation
Given:
A cyclist accelerates from 6 m/s at 1.5 m/s² for 5 s, then decelerates at 2 m/s² until its speed is 8 m/s . What is the total distance covered?
These values define the system as per NCERT data.
Formula:
Phase 1: v = 6 + 1.5 · 5 = 13.5 m/s, x_1 = 6 · 5 + 1/2 · 1.5 · (5)² = 30 + 18.75 = 48.75 m.
This is standard NCERT relation.
Substitution & Calculation:
Phase 2: 8 = 13.5 - 2 t Rightarrow t = 2.75 s, x_2 = 13.5 · 2.75 - 1/2 · 2 · (2.75)² = 37.125 - 7.5625 = 29.5625 m . Total = 48.75 + 29.5625 = 78.3125 m approx 78.31 m .
Result:
The computed value matches expected outcome and confirms correct choice as per NCERT.
Discussion
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