Question

At time $t=0\text{s}$, a particle starts from rest at position $(4\text{m},8\text{m})$. It starts moving towards the positive $x$-axis with a constant acceleration of $2{\text{m/s}}^2$. After $2\text{s}$, it undergoes an additional acceleration of $4{\text{m/s}}^2$ in the positive $y$-direction. Find the coordinates of the particle (in $\text{m}$) after the next $5\text{s}$.

• A. $(58,6)$
• B. $(53,58)$
• C. $(-53,58)$
• D. $(58,-6)$

Answer 1

The correct option is B $(53,58)$

Given,
Initial coordinates of particle $(x_0,y_0)=(4,8)$
Initial velocity of the particle $(u_x,u_y)=(0,0)$
From the data given in the question,
Particle is moving along $x$- axis with a constant acceleration $a_x=2{\text{m/s}}^2$
Therefore, we can use the kinematic formulae.
Using $v=u+at$, we can write that,
$v_x=u_x+a_{x}t$
At $t=2\text{sec}$ ,
$v_x=0+2\times 2\Rightarrow v_x=4\text{m/s}$
$v_y=0+0\times 2\Rightarrow v_y=0\text{m/s}$
Position of the particle at $t=2\text{s}$ is given by
$x_1=x_0+u_{x}t+\frac{1}{2}{a}_x{t}^2$
$=4+0+\frac{1}{2}\times \left(2\right)\times 2^2=8\text{m}$
$y_1=y_0+u_{y}t+\frac{1}{2}{a}_y{t}^2$
$=8+0+\frac{1}{2}\times \left(0\right)\times 2^2=8\text{m}$
$\Rightarrow$ Position of the particle at $t=2\text{s}$ is $(x_1,y_1)=(8,8)\text{m}$

Now, after next $5$ seconds,
Position of the particle is given by
$x_2=x_1+v_{x}t+\frac{1}{2}{a}_x{t}^2$
$y_2=y_1+v_{y}t+\frac{1}{2}{a}_y{t}^2$
From the given data ,
$x_2=8+\left(4\right)\times 5+\frac{1}{2}\times 2\times (5{)}^2=53\text{m}$
$y_2=8+\left(0\right)\times 5+\frac{1}{2}\times 4\times (5{)}^2=58\text{m}$
Hence, the position of the particle is given by $(x_2,y_2)=(53\text{m},58\text{m})$
Thus, option (b) is the correct answer.