Correct option is 3) 750 m
Initial velocity \((u)=0,\)
Acceleration \(\left(a_1\right)=2 m / s^2\) and time during acceleration \(\left(t_1\right)=10 \mathrm{sec}\).
Time during constant velocity \(\left(t_2\right)=30 \mathrm{sec}\) and retardation \(\left(a_2\right)=-4 m / s^2.\)
(- ve sign due to retardation).
Distance covered by the particle during acceleration,
\(s_1=u t_1+\frac{1}{2} a_1 t_1^2=(0 \times 10)+\frac{1}{2} \times 2 \times(10)^2 \)
\( =100 \mathrm{~m} \) ......(i)
And velocity of the particle at the end of acceleration,
\(v=u+a_1 t_1=0+(2 \times 10)=20 \mathrm{~m} / \mathrm{s}\)
Therefore distance covered by the particle during constant velocity \(\left(s_2\right)\)
\(=v \times t_2=20 \times 30=600 \mathrm{~m}\) ......(ii)
Relation for the distance covered by the particle during retardation \(\left(s_3\right)\) is
\(v^2=u_2+2 a_2 S_3\)
or, \((0)^2=(20)^2+2 \times(-4) \times s_3=400-8 s_3\)
or, \(s_3=400 / 8=50 \mathrm{~m} \) ......(iii)
Therefore total distance covered by the particle
\(s=s_1+s_2+s_1 \)
\(=100+600+50=750 \mathrm{~m}\).
