Question

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

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

Answer 1

Answer: (A)

From $\overrightarrow{F}=\left(10\hat{i}+5\hat{j}\right)\text{N}$ , we have
$F_x=10\text{ N}\text{; }F{}_y=5\text{ N}$
$a_x=\frac{F{}_x}{m}=\frac{10}{\text{0.01}}=1000{\text{ m s}}^{-2}$
As acceleration along the x-axis is constant,
$\therefore x=u{}_{x}t+\frac{1}{2}a{}_{x}t{}^2$
$=0+\frac{1}{2}\times 1000\times 5^2=12500\text{ m}$
Similarly, $a_y=\frac{F{}_y}{m}=\frac{5}{\text{0.01}}=500{\text{ m s}}^{-2}$
and $y=u{}_{y}t+\frac{1}{2}a{}_{y}t{}^2=0+\frac{1}{2}\times 500\times 5^2=6250\text{ m}$
Hence the position of the particle at $t=5s$ is
$\overrightarrow{r}=\left(12500\hat{i}+6250\hat{j}\right)\text{m}$