A particle starts from rest and has an acceleration of $2{\text{m/s}}^2$ for 10 sec. After that, it travels for 30 sec with constant speed and then undergoes a retardation of $4{\text{m/s}}^2$ and comes back to rest. The total distance covered by the particle is

The correct option is B $750\text{m}$
Given
Initially
$u=0\text{m/s};a=2{\text{m/s}}^2;t=10\text{s}$
Velocity after $10\text{s}$
Using $1^{st}$ equation of motion
$v=u+at$
$v=0+2\times 10=20\text{m/s}$
Distance covered in this time
Using $2^{nd}$ equation of motion
$s=ut+\frac{1}{2}a{t}^2$
$s_1=0+\frac{1}{2}\times 2\times {10}^2=100\text{m}$
After $10\text{s}$ velocity is $20\text{m/s}$ and it moves with this velocity for another $30\text{s}$
Distance covered during this $30\text{s}$
$s_2=20\times 30=600\text{m}$
At last it retards from velocity $20\text{m/s}$ to $0\text{m/s}$ with deceleration of $4{\text{m/s}}^2$
Distance covered during retardation
Using $3^{rd}$ equation of motion
$v^2=u^2+2as$
$0^2={20}^2-2\times 4\times s_3$
$s_3=50\text{m}$
Total distance travelled $=s_1+s_2+s_3=100+600+50=750\text{m}$