The correct option is B 4
Acceleration – time graph of the particle is a straight line as shown in the figure. Area enclosed by the a−t
graph and the x
-axis gives the change in velocity .
Particle will come to rest, when area above time axis and below becomes equal.
Area below time axis, A1=12×4×2=4
Area above time axis, A2=12×t×a=A1=4 ⇒at=8(1)
Slope of the line, m=42=2
Solving (1) and (2), we get t=2 s
Hence after t+2
i.e. 4 sec
, particle comes to rest.
Given, particle starts from rest i.e. at t=0,v=0
As we know that a=dvdt⇒ v(t)=∫t0adt ⇒ v(t)=∫t0(2t−4)dt ⇒ v(t)=t2−4t
The particle will come to rest only if v=0 ⇒ t2−4t=0⇒ t(t−4)=0 ⇒ t=0,4
Hence, after 4 sec
, particle will come to rest.