The distance between two particles is decreas | vedclass

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(A) Let the speeds of the two particles be $v_1$ and $v_2$ (where $v_1 > v_2$).When the particles move towards each other,their relative velocity is t

Vedclass · Accepted Answer

$5 \,m/s; 1 \,m/s$

Vedclass · Answer

(A) Let the speeds of the two particles be $v_1$ and $v_2$ (where $v_1 > v_2$).When the particles move towards each other,their relative velocity is the sum of their speeds. Given that the distance is decreasing at $6 \,m/s$,we have:$v_1 + v_2 = 6$ --- $(i)$When the particles move in the same direction,their relative velocity is the difference of their speeds. Given that the separation increases at $4 \,m/s$,we have:$v_1 - v_2 = 4$ --- (ii)Adding equations $(i)$ and (ii):$(v_1 + v_2) + (v_1 - v_2) = 6 + 4$$2v_1 = 10 \implies v_1 = 5 \,m/s$Substituting $v_1 = 5$ into equation $(i)$:$5 + v_2 = 6 \implies v_2 = 1 \,m/s$Thus,the speeds are $5 \,m/s$ and $1 \,m/s$.

The distance between two particles is decreasing at the rate of $6 \,m/s$. If these particles travel with the same speeds and in the same direction,then the separation increases at the rate of $4 \,m/s$. The speeds of the particles are:

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