Q. A particle leaves the origin with an initial velocity $\overrightarrow{\mathbf{v}} = \left( 3.00\widehat{\mathbf{i}} \right)ms^{- 1}$ and a constant acceleration $\overrightarrow{\mathbf{a}} = \left( - 1.00\widehat{\mathbf{i}} - 0.50\widehat{\mathbf{j}} \right)\ ms^{- 2}$ . When the particle reaches it maximum $x$-coordinate, what is its $y$-component a velocity?
A) $- 2.0\ ms^{- 1}$
B) $- 1.0\ ms^{- 1}$
C) $- 1.5\ ms^{- 1}$
D) $1.0\ ms^{- 1}$
Explanation
(c) The velocity of the particle at any time $t$
$\overrightarrow{\mathbf{v}} = \overrightarrow{\mathbf{v}}0 + \overrightarrow{\mathbf{a}}t$
The $x$-component is
$v_{x} = v_{ax} + a_{x}t$
The $y$-component is
$v_{y} = v_{ay} + a_{x} = ( - 0.5t)ms^{- 1}$
When the particle reaches its maximum $x$-coordinates, $v_{x} = 0$. That is
$3 - t = 0$
$\Rightarrow t = 3s$
The $y$-component of the velocity of this time is
$v_{y} = - 0.5 \times 3 = - 1.5\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.