The displacement x and time for a particle are related to each other as t = sqrt(x) + 3. What is work done in first six seconds of its motion?

The displacement x and time for a particle are related to each other a

1.	The displacement x and time for a particle are related to each other as t = sqrt(x) + 3. What is work done in first six seconds of its motion?
Answer» 6J ZERO 4J 2J Solution :From `sqrt(x) + 3 = t, x = (t - 3)^2` Now, `v = (DX)/(dt) = 2(t -3)` At `t = 0, v_1 = 2(-3) = -6` At `t = 6 , v_2 = 2(6 - 3) = 6` Work done = Change in KE = `1/2 m (v_2^2 - v_1^2)` = zero.