91Ó°ÊÓ

The coefficient of kinetic friction between tire and road is ${\mathrm{μ}}_K=0.65$

The limiting speed for car A is: $\mathrm{u}=35 \mathrm{mph}$

The skidding distance for both cars is $\mathrm{x}=7.15 \mathrm{m}$

The mass of block A is: ${\mathrm{m}}_{\mathrm{a}}=1500 \mathrm{kg}$

The mass of block B is: ${\mathrm{m}}_{\mathrm{b}}=1900 \mathrm{kg}$

The speed of both car cars after collision is found by using work energy theorem and then applies momentum conservation between cars to find speed of car A before collision.

Apply the work energy theorem to find the combined speed of cars.

$$\begin{gathered} \frac{1}{2}\left(m_a+m_b\right)v^2={\mathrm{μ}}_k\left(m_a+m_b\right)gx \\ v=\sqrt{2{\mathrm{μ}}_{k}gx} \end{gathered}$$

Here, $g$is the gravitational acceleration.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{v}=\sqrt{2\left(0.65\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(7.15\mathrm{m}\right)} \\ \mathrm{v}=9.54\mathrm{m}/\mathrm{s} \end{gathered}$$

Apply the conservation of momentum for car A and car B.

$m_a{u}_a+m_b{u}_b=(m_a+m_b)v$

Here,$u_a\mathrm{and}{u}_b$ is the initial speed of car A and car B.

Substitute all the values in the above equation.

$$\begin{gathered} \left(1500\mathrm{Kg}\right){\mathrm{u}}_{\mathrm{a}}+\left(1900\mathrm{Kg}\right)\left(0\right)=\left(1500\mathrm{Kg}+1900\mathrm{kg}\right)\left(9.54\mathrm{m}/\mathrm{s}\right) \\ {\mathrm{u}}_{\mathrm{a}}=21.62\mathrm{m}/\mathrm{s} \\ {\mathrm{u}}_{\mathrm{a}}=\left(21.62\mathrm{m}/\mathrm{s}\right)\left(\frac{1\mathrm{mi}}{1600\mathrm{m}}\right)\left(\frac{3600\mathrm{s}}{1\mathrm{hr}}\right) \\ {\mathrm{u}}_{\mathrm{a}}=48.64\mathrm{mph} \end{gathered}$$

Therefore, the speed of car A before collision is$48.64\mathrm{mph}$.

As the speed of car A before collision $\left(48.64\mathrm{mph}\right)$is more than the speed limit for car A, so the car A is speeding.

The exceeding speed of car A above speed limit is calculated as:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{e}}=\mathrm{v}-\mathrm{u} \\ {\mathrm{v}}_{\mathrm{e}}=48.64\mathrm{mph}-35\mathrm{mph} \\ {\mathrm{v}}_{\mathrm{e}}=13.64\mathrm{mph} \end{gathered}$$

Therefore, the car A is speeding above the speed limit by speed $13.64\mathrm{mph}$.