Question

A particle starts from the origin at $t=0s$ with a velocity of $10\hat{j}m{s}^{-1}$ and move in the $x-y$ plane with a constant acceleration of $(8\hat{j}+2\hat{j})m{s}^{-2}$. At an instant when the $x$-coordinate of the particle is $16m,y$-coordinate of the particle is:

• A. 16 m
• B. 28 m
• C. 36 m
• D. 24 m

Answer 1

Answer: (D)

$\vec{r}=\vec{u}t+\frac{1}{2}\vec{a}{t}^2$
$x\hat{i}+y\hat{j}=10t\hat{j}+\frac{1}{2}(8\hat{i}+2\hat{j})t^2$
$x\hat{i}+y\hat{j}=10t\hat{j}+(4\hat{i}+\hat{j})t^2$
$16\hat{i}+yj=10t\hat{j}+4t^2\hat{i}+t^2\hat{j}$
$16\hat{i}+\hat{j}=4t^2\hat{i}+\left(10t+t^2\right)\hat{j}$
$4t^2=16$
$t=4s$
$y=10t+t^2$
$=20+4=24m.$