Two cars get closer by 8,m every second while | vedclass

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(B) Let the speeds of the two cars be $v_1$ and $v_2$ (where $v_1 > v_2$).When travelling in opposite directions,the relative speed is the sum of thei

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$4.4\,ms^{-1}$ and $3.6\,ms^{-1}$

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(B) Let the speeds of the two cars be $v_1$ and $v_2$ (where $v_1 > v_2$).When travelling in opposite directions,the relative speed is the sum of their individual speeds:$v_1 + v_2 = \frac{8\,m}{1\,s} = 8\,ms^{-1}$ (Equation $1$)When travelling in the same direction,the relative speed is the difference of their individual speeds:$v_1 - v_2 = \frac{0.8\,m}{1\,s} = 0.8\,ms^{-1}$ (Equation $2$)Adding Equation $1$ and Equation $2$:$(v_1 + v_2) + (v_1 - v_2) = 8 + 0.8$$2v_1 = 8.8$$v_1 = 4.4\,ms^{-1}$Substituting $v_1$ in Equation $1$:$4.4 + v_2 = 8$$v_2 = 8 - 4.4 = 3.6\,ms^{-1}$Thus,the speeds of the two cars are $4.4\,ms^{-1}$ and $3.6\,ms^{-1}$.

Two cars get closer by $8\,m$ every second while travelling in the opposite directions. They get closer by $0.8\,m$ in every second while travelling in the same direction. What are the speeds of the two cars?

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