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Physics · MOTION IN A STRAIGHT LINE
हिन्दीA car starting from rest accelerates at a constant rate \( a_1 = 4.0 \text{ m/s}^2 \) for time \( t_1 \), and then decelerates at a constant rate \( a_2 = 6.0 \text{ m/s}^2 \) for time \( t_2 \) until it comes to a complete stop. If the total distance covered during this motion is \( S = 120 \text{ m} \), calculate the maximum velocity \( v_{\max} \) attained by the car.
Correct answer · Option 3
\( 24 \text{ m/s} \)
Explanation
Distance during acceleration \( s_1 = \frac{v_{\max}^2}{2 a_1} \), distance during deceleration \( s_2 = \frac{v_{\max}^2}{2 a_2} \). Total distance \( S = \frac{v_{\max}^2}{2} (\frac{1}{a_1} + \frac{1}{a_2}) = \frac{v_{\max}^2 (a_1 + a_2)}{2 a_1 a_2} \). Thus \( v_{\max} = \sqrt{\frac{2 S a_1 a_2}{a_1 + a_2}} = \sqrt{\frac{2 \times 120 \times 4 \times 6}{4 + 6}} = \sqrt{576} = 24 \text{ m/s} \).
