Two cars A and B initially at rest are moving | vedclass

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(B) Using the first equation of motion,$v = u + at$. Since both cars start from rest $(u = 0)$,their velocities after time $t$ are:$v_1 = a_1 t$ and $

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) Using the first equation of motion,$v = u + at$. Since both cars start from rest $(u = 0)$,their velocities after time $t$ are:$v_1 = a_1 t$ and $v_2 = a_2 t$Subtracting these,we get: $(v_1 - v_2) = (a_1 - a_2)t \dots (1)$Using the second equation of motion,$s = ut + \frac{1}{2}at^2$. The distance between them is:$s_1 - s_2 = \frac{1}{2}a_1 t^2 - \frac{1}{2}a_2 t^2 = \frac{1}{2}(a_1 - a_2)t^2$Given $s_1 - s_2 = 50 \ m$ and $(a_1 - a_2) = 4 \ m \ s^{-2}$,we have:$50 = \frac{1}{2} 	imes 4 	imes t^2 \Rightarrow 50 = 2t^2 \Rightarrow t^2 = 25 \Rightarrow t = 5 \ s$Substituting $t = 5 \ s$ and $(a_1 - a_2) = 4 \ m \ s^{-2}$ into equation $(1)$:$(v_1 - v_2) = 4 	imes 5 = 20 \ m \ s^{-1}$.

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