A particle moving with a velocity equal to 0.4 m/s is subjected to an uniform acceleration of \( 0.15 m/s^{2} \) for 2 sec in a direction at right angles to its initial direction of motion. The resultant velocity is :
Solution:
Initial velocity along x-axis is \(v_x = 0.4\text{ m/s}\). Velocity developed along the perpendicular y-axis is \(v_y = a_y t = 0.15 \times 2 = 0.3\text{ m/s}\). Resultant velocity is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{0.4^2 + 0.3^2} = 0.5\text{ m/s}\).
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