First slide

Rectilinear Motion

Question

A particle starts from the origin with a velocity of 10 m s^-1 and moves with a constant acceleration till the velocity increases to 50 ms^-1. At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?

Moderate

A

Zero
B

$10 {ms}^{- 1}$
C

$50 {ms}^{- 1}$
D

$70 {ms}^{- 1}$

Solution

$\begin{array}{l} 2 a x = (50)^{2} - (10)^{2} and 2 (- a) (- x) = v^{2} - (50)^{2} \\ This gives v^{2} - (50)^{2} = (50)^{2} - (10)^{2} i.e. v = 70 {ms}^{- 1} \end{array}$

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