Question

The force on a particle of mass $10\text{g}$ is $(10\hat{i}+5\hat{j})$. If it starts from rest from origin, what would be its position at time $\text{t}=5\text{s}?$

• A. $(6250\hat{i}+12500\hat{j})\text{m}$
• B. $(-12500\hat{i}+6250\hat{j})\text{m}$
• C. $(12500\hat{i}-6250\hat{j})\text{m}$
• D. $(12500\hat{i}+6250\hat{j})\text{m}$

Answer 1

The correct option is D $(12500\hat{i}+6250\hat{j})\text{m}$

We have $F_x=10\text{N}$ giving
$a_x=\frac{F_x}{m}=\frac{10\text{N}}{0.01\text{kg}}=1000{\text{m/s}}^2$
As this is a case of constant acceleration in $x-$direction,
$x=u_{x}t+\frac{1}{2}{a}_x{t}^2=\frac{1}{2}\times 1000{\text{m/s}}^2\times {\text{(5 s)}}^2$
$x=12500\text{m}$
Similarly, $a_y=\frac{F_y}{m}=\frac{5\text{N}}{0.01\text{kg}}=500{\text{m/s}}^2$
and $y=6250\text{m}$
Thus, the position of the particle at $t=5\text{s}$ is,
$\overrightarrow{r}=(12500\hat{i}+6250\hat{j})\text{m}$