Question

A particle is moving in $xy-$plane with $y=\frac{x}{2}$ and velocity along $x$ direction is $u_x=4-2t$. The correct options are

• A. Initial velocities of $x$ and $y$ directions are negative.
• B. Initial velocities of $x$ and $y$ directions are positive.
• C. Motion is first retarded, then accelerated
• D. Motion is first accelerated, then retarded.

Answer 1

Answer: (B), (C)

The correct options are
B Initial velocities of $x$ and $y$ directions are positive.
C Motion is first retarded, then accelerated


$v_x=4-2t,$
Comparing with equation of kinematics
$v_x=u_x+a_{x}t\Rightarrow u_x=4,a_x=-2$

Now, $y=\frac{x}{2}\Rightarrow \frac{dy}{dt}=\frac{1}{2}\frac{dx}{dt}$
$v_y=\frac{1}{2}{v}_x=2-t,$
Comparing with equation of kinematics
$v_y=u_y+a_{y}t,$
$u_y=2and{a}_y=-1$
Initial velocities in both $x,y$ directions are positive, acceleration in both $x,y$ direction is negative.
So motion retards initially.
After $t=2\text{s},$ velocity in both $x,y$ directions become negative, particle accelerates.