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Physics · MOTION IN A STRAIGHT LINE
हिन्दीA train accelerates uniformly from rest at rate \( a_1 = 1.0 \text{ m/s}^2 \) for a time \( t_1 \), then travels at a constant maximum velocity \( v_{\max} \) for a time interval \( t_2 = 10 \text{ s} \), and finally decelerates uniformly at rate \( a_2 = 2.0 \text{ m/s}^2 \) until coming to rest. If the total distance covered is \( 500 \text{ m} \), find \( v_{\max} \).
Correct answer · Option 1
\( 20 \text{ m/s} \)
Explanation
Distances: \( s_1 = \frac{v_{\max}^2}{2 a_1} = \frac{v_{\max}^2}{2} \), \( s_2 = v_{\max} t_2 = 10 v_{\max} \), \( s_3 = \frac{v_{\max}^2}{2 a_2} = \frac{v_{\max}^2}{4} \). Total distance \( S = \frac{3}{4} v_{\max}^2 + 10 v_{\max} = 500 \implies 3 v_{\max}^2 + 40 v_{\max} - 2000 = 0 \). Solving gives \( (v_{\max} - 20)(3 v_{\max} + 100) = 0 \implies v_{\max} = 20 \text{ m/s} \).
