Starting from rest a particle moves in a straight line with acceleration $a = \{2 + |t – 2|\} m/s^2$. Velocity of particle at the end of 4 s will be

12 m/s

The correct answer is Option (4) → 12 m/s

Acceleration can be written as $a = 2 + 2 – t$ or $a = 4 – t$ for $t ≤ 2s$ and $a = 2 + t – 2$ or $a = t$ for $t ≥ 2 s$

Therefore, acceleration time graph of the particle will be as shown below

Now since, $dv = a dt$

$v_f − v_i$ = area under (a-t) graph

or $v_f −0=(4 × 4)-\frac{1}{2}(4)(2)$

$=12 m/s$

or velocity of particle at the end of 4s is 12 m/s.