On a body of 2 kg mass at rest a force of 10 N is applied. What is the distance travelled by the body in 2 sec?

Given:
$\mathrm{Mass} =\mathrm{ }2 \mathrm{kg}$
$\mathrm{Time}, \mathrm{t}=2 \sec$
$\mathrm{Distance} = \mathrm{S}$
$\mathrm{Initial} velocity, \mathrm{u} =0\mathrm{ }$

F = 10 N
By using Newton’s second law, we have $F=\mathit{ma}$
Therefore, $a=\frac{F}{m}$
$\Rightarrow \mathit{ a}=\frac{10}{2}$
$\Rightarrow \mathrm{a}=5 \mathrm{m}/{\mathrm{s}}^2$   ----- (1)
We know,  $\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ 

Where displacement $\left(s\right)$of an object, velocity $\left(u\right)$ time$\left(t\right).$
$\Rightarrow \mathit{ S}= 0+\frac{1}{2} \times 5 \times 2^2$
$\Rightarrow \mathit{ S}=10\mathit{ m }$
Therefore, Distance travelled in $2 \sec$ = 10 m.