What are the speeds of two objects if,when th | vedclass

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(D) Let the speeds of the two objects be $v_A$ and $v_B$ (where $v_A > v_B$).When they move towards each other,their relative speed is the sum of thei

Vedclass · Accepted Answer

$2.2\,m/s$ and $1.8\,m/s$

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(D) Let the speeds of the two objects be $v_A$ and $v_B$ (where $v_A > v_B$).When they move towards each other,their relative speed is the sum of their individual speeds: $v_{rel} = v_A + v_B$.Given that they get $4\,m$ closer in $1\,s$,the relative speed is $v_A + v_B = \frac{4\,m}{1\,s} = 4\,m/s$ (Equation $1$).When they move in the same direction,their relative speed is the difference of their individual speeds: $v_{rel} = v_A - v_B$.Given that they get $4.0\,m$ closer in $10\,s$,the relative speed is $v_A - v_B = \frac{4.0\,m}{10\,s} = 0.4\,m/s$ (Equation $2$).Adding Equation $1$ and Equation $2$: $(v_A + v_B) + (v_A - v_B) = 4 + 0.4 \implies 2v_A = 4.4 \implies v_A = 2.2\,m/s$.Subtracting Equation $2$ from Equation $1$: $(v_A + v_B) - (v_A - v_B) = 4 - 0.4 \implies 2v_B = 3.6 \implies v_B = 1.8\,m/s$.Thus,the speeds are $2.2\,m/s$ and $1.8\,m/s$.

What are the speeds of two objects if,when they move uniformly towards each other,they get $4\,m$ closer in each second and when they move uniformly in the same direction with the original speeds,they get $4.0\,m$ closer each $10\,s$?

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