newcommand{tx}[1]{text{#1}}

subsection*{Known}

begin{align}
&v_{av}=frac{v_i+v_f}{2}notag\
&implies x_f=x_i+left(frac{v_i+v_f}{2}right)t, tx{for} a=tx{const}.
end{align}
subsection*{Calculation}
Givens: $x_{f_{tx{police}}}=x_{f_{tx{speeder}}}=x_fimplies t_{tx{police}}=t_{tx{speeder}}=t$ (or conversely), $x_{i_{tx{police}}}=x_{i_{tx{speeder}}}=0 tx{m}$, $v_{i_{tx{police}}}=0 frac{tx{m}}{tx{s}}$, $v_{i_{tx{speeder}}}=v_{f_{tx{speeder}}}$.\

From (1) we have:
begin{align*}
&x_f=left(frac{v_{f_{tx{police}}}}{2}right)t=left(frac{v_{i_{tx{speeder}}}+v_{f_{tx{speeder}}}}{2}right)t\
&implies frac{v_{f_{tx{police}}}}{2}=frac{cancel{2}v_{f_{tx{speeder}}}}{cancel{2}}\
&thereforeboxed{v_{f_{tx{police}}}=2v_{f_{tx{speeder}}}}
end{align*}

vspace{2pt}
hrule
vspace{1pt}

subsection*{Conclusion}

begin{align*}
boxed{v_{f_{tx{police}}}=2v_{f_{tx{speeder}}}}
end{align*}

$$
begin{align*}
boxed{v_{f_{text{police}}}=2v_{f_{text{speeder}}}}
end{align*}
$$

$$
textbf{Concept:}
$$

From the data the velocity of the speeder is known to be18.2 m/s whereas the police car accelerates at 4.50 $m/s^2$, it catches up with the speeder at $t=8.09s$. Hence if the police car accelerates at $5.0 m/s^2$, we expect the police car to catch up at an earlier time. From the results of last question we can find the time at which the police car catches up: the two cars will meet when the final velocity of the police car is twice the velocity of the speeder, or 36.4 m/s.

$$
textbf{Solution:}
$$

As 5.00 $m/s^2$ is a faster acceleration as compared to $4.50 m/s^2$, we expect the police car to catch up with the speeder at a time that is less than $t=8.09s$.
Assume the final velocity of the police car to be two times the velocity of the speeder (36.4 m/s) and use the velocity-time equation to find the time the two cars will meet

$$
v_f=v_i+at Rightarrow t=frac{v_f-v_i}{a}=frac{36.4-0m/s}{5.0m/s^2}=color{#4257b2} boxed{bf 7.28s}
$$

$$
t=7.28s
$$

In this problem, at time $t = 0$, car 1 passes through marker 0 ($x_text{i, 1} = 0$) traveling with speed $s_text{i, 1} = 20.0~mathrm{m/s}$ due east. At the same time, car 2 is at $x_text{i, 2} = 1000~mathrm{m}$ east of marker 0 with speed $s_text{i, 2} = 30.0~mathrm{m/s}$. Car 1 is speeding with at rate $2.5~mathrm{m/s^{2}}$, while car 2 is slowing down at rate $-3.20~mathrm{m/s}$. We write the position-time equations for the two cars, and find the time in which they meet. Let eastward be the positive direction.

Part A.

Car 1 is initially moving due east, so the velocity must be positive $v_text{i, 1} = +20.0~mathrm{m/s}$. It is speeding up, so the acceleration and velocities are in the same direction. Its acceleration must also be positive, hence $a_{1} = +2.5~mathrm{m/s^{2}}$. The position-time equation must be

$$
begin{align*}
x_text{f, 1} &= x_text{i, 1} + v_text{i, 1}t + frac{1}{2}a_{1}t^{2} \
x_text{f, 1} &= boxed{ left( +20.0~mathrm{m/s} right)t + frac{1}{2} left( +2.5~mathrm{m/s^{2}} right)t^{2} }
end{align*}
$$

Car 2 initially moving due west, so the velocity must be negative $v_text{i, 2} = -30.0~mathrm{m/s}$. It is slowing down, so the acceleration and velocities are in the opposite direction. Its acceleration must be positive, hence $a_{2} = +3.2~mathrm{m/s^{2}}$. The position-time equation must be

$$
begin{align*}
x_text{f, 2} &= x_text{i, 2} + v_text{i, 2}t + frac{1}{2}a_{2}t^{2} \
x_text{f, 2} &= boxed{ 1000~mathrm{m} + left( -30.0~mathrm{m/s} right)t + frac{1}{2} left( +3.2~mathrm{m/s^{2}} right)t^{2} }
end{align*}
$$

Part B.

The two cars meet when $x_text{f, 1} = x_text{f, 2}$. We have

$$
begin{align*}
x_text{f, 1} &= x_text{f, 2} \
left( +20.0~mathrm{m/s} right)t + frac{1}{2} left( +2.5~mathrm{m/s^{2}} right)t^{2} &= 1000~mathrm{m} + left( -30.0~mathrm{m/s} right)t + frac{1}{2} left( +3.2~mathrm{m/s^{2}} right)t^{2} \
0 &= 1000~mathrm{m} + left( -30.0~mathrm{m/s} – 20.0~mathrm{m/s} right)t + frac{1}{2} left( +3.2~mathrm{m/s^{2}} – 2.5~mathrm{m/s^{2}} right)t^{2} \
0 &= 1000~mathrm{m} + left( -50.0~mathrm{m/s} right)t + frac{1}{2} left( +0.70~mathrm{m/s^{2}} right)t^{2} qquad (1)
end{align*}
$$

Equation (1) is a quadratic, with roots

$$
begin{align*}
t &= -frac{-50.0~mathrm{m/s}}{+0.70~mathrm{m/s^{2}}} pm frac{sqrt{ left( -50.0~mathrm{m/s} right)^2 – 4 left( +0.70~mathrm{m/s^{2}} right) left( 1000~mathrm{m} right) }}{+0.70~mathrm{m/s^{2}}} \
t &= 24.04822~mathrm{s},~118.80893~mathrm{s}
end{align*}
$$

We take the smaller root. The larger root happens when the direction of car 2’s velocity has already flipped. Hence, the time that they meet is
$$
boxed{ t = 24.0~mathrm{s} }
$$

(a)
$$
x_text{f, 1} = left( +20.0~mathrm{m/s} right)t + frac{1}{2} left( +2.5~mathrm{m/s^{2}} right)t^{2}
$$

$$
x_text{f, 2} = 1000~mathrm{m} + left( -30.0~mathrm{m/s} right)t + frac{1}{2} left( +3.2~mathrm{m/s^{2}} right)t^{2}
$$

(b)
$$
t=24.0mathrm{,s}
$$