Two cars, at a certain instant, are 50 km ap | vedclass

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(B) Let car $A$ be at the origin $(0, 0)$ moving north with velocity $\vec{V}_A = 25 \hat{j} \ km/hr$. Let car $B$ be at $(0, 50)$ moving west with ve

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) Let car $A$ be at the origin $(0, 0)$ moving north with velocity $\vec{V}_A = 25 \hat{j} \ km/hr$. Let car $B$ be at $(0, 50)$ moving west with velocity $\vec{V}_B = -25 \hat{i} \ km/hr$.The relative velocity of $A$ with respect to $B$ is $\vec{V}_{AB} = \vec{V}_A - \vec{V}_B = 25 \hat{j} - (-25 \hat{i}) = 25 \hat{i} + 25 \hat{j} \ km/hr$.The magnitude of relative velocity is $|\vec{V}_{AB}| = \sqrt{25^2 + 25^2} = 25\sqrt{2} \ km/hr$.The relative position vector is $\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (0 - 0)\hat{i} + (0 - 50)\hat{j} = -50 \hat{j} \ km$.The time taken to reach the closest approach is given by $t = -\frac{\vec{r}_{AB} \cdot \vec{V}_{AB}}{|\vec{V}_{AB}|^2}$.$t = -\frac{(-50 \hat{j}) \cdot (25 \hat{i} + 25 \hat{j})}{(25\sqrt{2})^2} = -\frac{-1250}{1250} = 1 \ hr$.Thus, the time taken is $60 \ min$.

Column-$I$	Column-$II$
$(1)$ Velocity of $A$ relative to $B$ if $A$ and $B$ are moving perpendicular to each other	$(a)$ $\vec{v}_{rm} = \vec{v}_r + \vec{v}_m$
$(2)$ Velocity of rain drops relative to a man	$(b)$ $\vec{v}_{AB} = \vec{v}_A + \vec{v}_B$
	$(c)$ $v_{AB} = \sqrt{v_A^2 + v_B^2}$

Two cars, at a certain instant, are $50 \ km$ apart on a line running from south to north. The one farther north is moving west at $25 \ km/hr$. The other is moving towards north at $25 \ km/hr$. How long do they take to reach their distance of closest approach (in $min$)?

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