A particle starts from the origin with a velocity of 10 m/s and moves with a constant acceleration till the velocity increases to 50 m/s. At that instant the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?

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Solution

$Initial velocit y, u = 10 m s^{- 1} Final velocity, v = 50 m s^{- 1} From the third kinematics equation v^{2} - u^{2} = 2 a S \Rightarrow S = \frac{{(50)}^{2} - {(10)}^{2}}{2 a} = \frac{1200}{a} When the acceleration get reversed, then v^{2} - u^{2} = 2 a S \Rightarrow v^{2} = u^{2} + 2 (- a) (\frac{1200}{- a}) \Rightarrow v^{2} = {(50)}^{2} + 2400 = 4900 \Rightarrow v = 70 m s^{- 1}$

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A particle starts from the origin with a velocity of 10 m/s and moves with a constant acceleration till the velocity increases to 50 m/s. At that instant the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?

Initial velocity, u=10 m s-1Final velocity, v=50 m s-1From the third kinematics equationv2-u2=2aS⇒S=502-1022a=1200aWhen the acceleration get reversed, thenv2-u2=2aS⇒v2=u2+2-a1200-a⇒v2=502+2400=4900⇒v=70 m s-1