The correct option is B At t=7 s
From t=0 to t=4 s
Area under the a−t graph is given by
A1=−(1.5×4)=−6 m/s
Let the particle attain initial velocity again after time t′
So, from t=4 s to t=t′ s
Area under the a−t graph is given by
A2=2t′
We know that area under the a−t graph represents change in velocity.
As the initial and final velocities are same,
Δv=0 (where Δv is change in velocity)
⇒A1+A2=0
⇒−6+2t′=0
⇒t′=3 s
So, the desired time =4+3=7 s
Hence, the correct answer is option (b)