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Two people $A,~B$ are participating a running race.Initially of course they are both at rest.They then proceed with constant acceleration. $A$ covers the last $\dfrac{1}{4}$ of the distance in $3$ second while $B$ covers the last $\dfrac{1}{3}$ of his distance in $4$ second.

How long after the winner finishing the race will the other complete the race, assuming he completes the race?

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Let $a$ and $b$ be the accelerations of $A$ and $B$, respectively, and let $d$ be the length of the race. Let $x_A(t)$ and $x_B(t)$ be the distances covered by $A$ and $B$, respectively, at time $t$ after the start of the race; $x_A(t)=\frac12at^2$, and $x_B(t)=\frac12bt^2$. $A$ reaches the $3/4$ mark at time $t_1$ given by the equation $$\frac34d=\frac12at_1^2\tag{1}$$ and finishes the race at time $t_2$ given by the equation $$d=\frac12at_2^2\;,\tag{2}$$ and we’re told that $t_2-t_1=3$, so we can rewrite $(2)$ as $$d=\frac12a(t_1+3)^2\tag{3}$$ and solve $(1)$ and $(3)$ for $t_1$. Multiplying $(1)$ by $4/3$, we get $$d=\frac23at_1^2\;,$$ so

$$\begin{align*} &\frac23at_1^2=\frac12a(t_1+3)^2\;,\\ &4t_1^2=3(t_1+3)^2=3t_1^2+18t_1+27\;,\\ &t_1^2-18t_1-27=0\;,\\ &t_1=\frac12(18\pm\sqrt{432})=9\pm 6\sqrt3\;, \end{align*}$$

and since $3-6\sqrt3<0$, we must have $t_1=9+6\sqrt3$ and $t_2=12+6\sqrt3$.

Following the same procedure for $B$, starting with the system

$$\left\{\begin{align*} &\frac23d=\frac12bt_3^2\\ &d=\frac12b(t_3+4)^2\;, \end{align*}\right.$$

we get $3t_3^2=2(t_3+4)^2=2t_3^2+16t_3+32$, $t_3^2-16t_3-32=0$, $t_3=8+4\sqrt6$, and $t_3+4=12+4\sqrt6$.

Thus, $A$ finishes at time $12+6\sqrt3\approx 22.392$, and $B$ finishes at time $12+4\sqrt6\approx 21.798$. $B$ is the winner by $6\sqrt3-4\sqrt6=(6-4\sqrt2)\sqrt3\approx 0.343$ seconds.

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Choose the unit of length so that the distance raced is $1$. Let $a$ and $b$ be the accelerations of A and B, and let $t_a$ and $t_b$ be the times taken by A and B.

The distance travelled (from rest) in time $t$ if the acceleration is $k$ is $(1/2)kt^2$. Thus $$at_a^2 =2 \qquad \text{and}\qquad bt_b^2=2.\qquad \qquad(\ast)$$

Runner A travels the last $1/4$ of the distance in $3$ seconds. In the last $3$ seconds of her run, the distance travelled by A is $(1/2)(a)(t_a^2-(t_a-3)^2)$. Similarly, the distance travelled by B in the last $4$ seconds of her run is $(1/2)(b)(t_b^2-(t_b-4)^2)$. This yields the equations $$a(t_a^2-(t_a-3)^2)=\frac{2}{4}\qquad\text{and}\qquad b(t_b^2-(t_b-4)^2)=\frac{2}{3}.\qquad\qquad(\ast\ast)$$

To solve, divide, using the first equation of $(\ast)$ and of $(\ast\ast)$, and simplify slightly. We get $$\frac{t_a^2}{6t_a-9}=4.$$ We get a similar equation for $t_b$. Each equation simplifies to a quadratic.

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