Question

A particle starts from the origin at $t=0$ with an initial velocity of

$\begin{array}{l}\overrightarrow{u}=3\hat{i}\end{array}$

from origin and moves in the $x-y$ plane with a constant acceleration

$\begin{array}{l}\overrightarrow{a}=6\hat{i}+4\hat{j}m/s^2\end{array}$.

The $x-$coordinate of the particle at the instant when its $y-$coordinated is $32$ meters. The value of $D$ is:

• A. $60$
• B. $50$
• C. $32$
• D. $40$

Answer 1

The correct option is A

$60$

Step 1. Here acceleration is constant so we can use equations of motion.

Now taking motion along the Y-axis:

$\Rightarrow$ Displacement along the y-axis$(s_y)=32m$

$\Rightarrow$ Acceleration along the y-axis $(a_y)=4m{s}^{-2}$

$\Rightarrow$ Initial velocity along the y-axis $(u_y)=0$

using the second equation of motion

$\begin{array}{l}{S}_y=u_{y}t+\frac{1}{5}{a}_y{t}^2\end{array}$

$\begin{array}{l}\Rightarrow 32=0+\frac{1}{2}4t^2\end{array}$

$\Rightarrow t^2=16$

$\Rightarrow t=4sec$

Step 2. Now taking motion along the x-axis:

$\Rightarrow$ Displacement along x-axis $(s_x)=?$

$\Rightarrow$ Acceleration along x-axis $\left(a_x\right)=6$

$\Rightarrow$ Initial velocity along the x-axis $(u_x)=3$

$\begin{array}{l}{S}_x=u_{x}t+\frac{1}{5}{a}_x{t}^2\end{array}$

$$\begin{gathered} \Rightarrow S_x=3x4+\left(\frac{1}{2}\right)x6x16 \\ \therefore S_x=60m \end{gathered}$$

Hence, the correct option is $\left(A\right)$.