An iron rod of length 1.5 ,m lying on a horiz | vedclass

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(A) Let the rod be $AB$ with length $l = 1.5 \,m$. Let the end $A$ be pulled vertically upwards with constant velocity $v_A = 3 \,m/s$. Let $y$ be the

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{2}} \,s$

Vedclass · Answer

(A) Let the rod be $AB$ with length $l = 1.5 \,m$. Let the end $A$ be pulled vertically upwards with constant velocity $v_A = 3 \,m/s$. Let $y$ be the vertical height of end $A$ at time $t$, so $y = v_A t = 3t$. Let $x$ be the horizontal distance of end $B$ from the point directly below $A$ at time $t$. The rod maintains its length $l$, so by the Pythagorean theorem: $x^2 + y^2 = l^2$. Substituting $y = 3t$ and $l = 1.5$, we get $x^2 + (3t)^2 = (1.5)^2$, which simplifies to $x^2 + 9t^2 = 2.25$. Differentiating with respect to time $t$: $2x \frac{dx}{dt} + 18t = 0$. The speed of end $B$ is $v_B = |\frac{dx}{dt}| = \frac{18t}{2x} = \frac{9t}{x}$. We want the time $t$ when $v_B = v_A = 3 \,m/s$. So, $\frac{9t}{x} = 3 \Rightarrow x = 3t$. Substituting $x = 3t$ into the constraint equation $x^2 + 9t^2 = 2.25$: $(3t)^2 + 9t^2 = 2.25 \Rightarrow 9t^2 + 9t^2 = 2.25 \Rightarrow 18t^2 = 2.25 \Rightarrow t^2 = \frac{2.25}{18} = \frac{225}{1800} = \frac{1}{8}$. Thus, $t = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}} \,s$.

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