In the figure shown, blocks A and C start from rest and move to the right with acceleration aA=12 t m / s 2 and aC=3 m / s 2 . Here, t is in seconds. The time (in s) when block B again comes to rest is

Question

In the figure shown, blocks A and C start from rest and move to the right with acceleration aA=12 t m / s 2 and aC=3 m / s 2 . Here, t is in seconds. The time (in s) when block B again comes to rest is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2 aB=(3-12 t) 2((d vB/d t))=(3-12 t) 2 ∫ limits00 d vB=3 t-12 (t2/2) 0=t(3-6 t) ⇒ t=(1/2) s

Q. In the figure shown, blocks $A$ and $C$ start from rest and move to the right with acceleration $a_{A} = 12 t m / s^{2}$ and $a_{C} = 3 m / s^{2} .$ Here, $t$ is in seconds. The time (in $s$ ) when block $B$ again comes to rest is

$2 a_{B} = (3 - 12 t)$
$2 (\frac{d v _{B}}{d t}) = (3 - 12 t)$
$2 0 \int 0 d v_{B} = 3 t - 12 \frac{t ^{2}}{2}$
$0 = t (3 - 6 t)$
$\Rightarrow t = \frac{1}{2} s$

Q. In the figure shown, blocks A and C start from rest and move to the right with acceleration aA​=12tm/s2 and aC​=3m/s2. Here, t is in seconds. The time (in s) when block B again comes to rest is

2aB​=(3−12t) 2(dtdvB​​)=(3−12t) 20∫0​dvB​=3t−122t2​ 0=t(3−6t) ⇒t=21​s

Q. In the figure shown, blocks $A$ and $C$ start from rest and move to the right with acceleration $a_{A} = 12 t m / s^{2}$ and $a_{C} = 3 m / s^{2} .$ Here, $t$ is in seconds. The time (in $s$ ) when block $B$ again comes to rest is

$2 a_{B} = (3 - 12 t)$
$2 (\frac{d v _{B}}{d t}) = (3 - 12 t)$
$2 0 \int 0 d v_{B} = 3 t - 12 \frac{t ^{2}}{2}$
$0 = t (3 - 6 t)$
$\Rightarrow t = \frac{1}{2} s$