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4-1.Newton's Laws of Motion
normal
If acceleration of $A$ is $2 \,\,m/s^2$ to left and acceleration of $B$ is $1\,\,m/s^2$ to left, then acceleration of $C$ is
A$1 \,\,m/s^2$ upwards
B$1 \,\,m/s^2$ downwards
C$2 \,\,m/s^2$ downwards
D$2 \,\,m/s^2$ upwards
Solution
$l_{1}+l_{2}+l_{3}+2 l_{4}=$ constant$\frac{d l_{1}}{d t}+\frac{d l_{2}}{d t}+\frac{d l_{3}}{d t}+2 \frac{d l_{4}}{d t}=0$
$\frac{d^{2} l_{1}}{d t^{2}}+\frac{d^{2} l_{2}}{d t^{2}}+\frac{d^{2} l_{3}}{d t^{2}}+2 \frac{d^{2} l_{4}}{d t^{2}}=0$
$2+(2-1)-1+2 \frac{d^{2} 1_{4}}{d t^{2}}=0$
$a_{B}=-1 m / s^{2}$
$a_{B}=1 m / s^{2} \uparrow$
Standard 11
Physics
