The correct option is B 2 s
Area under a−t curve gives change in velocity.
From t=0 to t=2 seconds
Area under the a−t curve =−10×2=−20 m/s
⇒Δv=v2−v0=−20 m/s
⇒v2=−20 m/s ( velocity is in negative direction)
So, the body is speeding up from t=0 to t=2 s
From t=2 to t=4 seconds
Area under the the a−t curve = 10×2=20 m/s
⇒Δv=v4−v2=20 m/s
⇒v4=0
So, the body is speeding down from t=2 to t=4 seconds
Hence, the maximum velocity of the particle is at t=2 s.