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Physics · MOTION IN A STRAIGHT LINE
हिन्दीThe acceleration of a particle moving along the x-axis is given as a function of position by \( a(x) = \beta x + \alpha \), where \( \beta = 3.0 \text{ s}^{-2} \) and \( \alpha = 2.0 \text{ m/s}^2 \). If the particle starts at \( x = 0 \) with an initial velocity \( v_0 = 4.0 \text{ m/s} \), find its velocity when it reaches position \( x = 2.0 \text{ m} \).
Correct answer · Option 2
\( 6.0 \text{ m/s} \)
Explanation
Using \( a = v \frac{dv}{dx} \implies \int_{4}^v v \, dv = \int_0^2 (3x + 2) \, dx \). Integrating yields \( \frac{v^2 - 16}{2} = [ \frac{3}{2}x^2 + 2x ]_0^2 = 6 + 4 = 10 \). Thus \( v^2 - 16 = 20 \implies v^2 = 36 \implies v = 6.0 \text{ m/s} \).
