A police officer is investigating the scene of an accident where two cars collided at an intersection. One car with a mass of \(1100 \mathrm{kg}\) moving west had collided with a \(1300-\mathrm{kg}\) car moving north. The two cars, stuck together, skid at an angle of \(30^{\circ}\) north of west for a distance of \(17 \mathrm{m} .\) The coefficient of kinetic friction between the tires and the road is \(0.80 .\) The speed limit for each car was $70 \mathrm{km} / \mathrm{h} .$ Was either car speeding?

To find the force of friction acting on the two cars, we'll use the formula: $$F_{friction} = μ * F_{normal} $$ Since the normal force is equal to the total weight of the two cars, and because weight = mass × gravity, we can find the normal force. Let's calculate the force of friction: \(F_{normal} = (1100 + 1300) * 9.81\) Now, calculate the force of friction using the coefficient of kinetic friction: \(F_{friction} = 0.80 * F_{normal}\).

Using Newton's second law of motion, we can calculate the acceleration of the system: $$F = m * a$$ The force acting on the two cars is the force of friction in the negative direction. The total mass of the cars is \((1100+1300)\,\text{kg}\). Rearrange the equation as: $$a = \frac{-F_{friction}}{m_{total}}$$ Calculate the acceleration using the force of friction found in step 1: \(a = \frac{-F_{friction}}{(1100+1300)}\)

Now we need to determine the final combined velocity of the two cars to analyze the collision. We can use the formula: $$v^2 = u^2 + 2 * a * s$$ Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance covered. Since the cars came to a stop after skidding, \(v=0\) Rearrange the equation to find the initial velocity: $$u=\sqrt{- 2 * a * s}$$ We'll now calculate the initial velocity using the distance \(s=17\,\text{m}\) and acceleration found in step 2: \(u = \sqrt{- 2 * a * 17}\) The direction of the velocity is \(30^\circ\) north of west.

Now we will use conservation of momentum to determine the initial velocities of the individual cars: $$m_1\vec{v}_1 + m_2\vec{v}_2 = (m_1 + m_2)\vec{v_f}$$ Let the initial velocities of the two cars be \(\vec{v}_{1i}\) and \(\vec{v}_{2i}\). We know the final velocity \(\vec{v}_f\) and their masses \(m_1\) and \(m_2\). Separate the equation into two components (north and west): $$\begin{cases} 1100 * v_{1i_{West}} + 0 = (1100 + 1300) * v_f * \cos(30^\circ) \\ 0 + 1300 * v_{2i_{North}} = (1100 + 1300) * v_f * \sin(30^\circ) \end{cases}$$ Solve the system of equations for \(v_{1i_{West}}\) and \(v_{2i_{North}}\): $$\begin{cases} v_{1i_{West}} = (1100 + 1300) * v_f * \cos(30^\circ) /1100 \\ v_{2i_{North}} = (1100 + 1300) * v_f * \sin(30^\circ) /1300 \end{cases}$$ Calculate the initial velocities using the final velocity found in step 3: \(v_{1i_{West}} = (1100 + 1300) * u * \cos(30^\circ) /1100\) \(v_{2i_{North}} = (1100 + 1300) * u * \sin(30^\circ) /1300\)

Convert the speed limit from km/h to m/s: \(70\,\text{km/h} = 70 * (1000/3600) \,\text{m/s}\) Now compare the initial velocities \(v_{1i_{West}}\) and \(v_{2i_{North}}\) with the speed limit. If either of the initial velocities is greater than the speed limit, the respective car was speeding.