Answer:
Let the Volvo's speed upon braking be v, and its mass be m1.
The speed of the Volvo just before the crash is represented as v1.
After applying the brakes, the skidding distance is noted as 30 m.
The equation governing this scenario is v1^2- v^2 = - 2 * a * s, where a= represents deceleration and s= denotes skidding distance.
Assuming the deceleration to be -0.3 g = - 2.94 m/s^2 (as provided by the Volvo website for the car's emergency system).
The equation can be modified to solve for v: v^2 = v1^2 + (2 * a * s )= v1^2 + 176.4.
After the collision, both vehicles merge and proceed together (m1+m2) at an initial speed of V (let's say).
They then move 12.25 m and eventually come to a stop.
Using the equation (0)^2 -(V)^2 = - 2 * a' * s', where a' signifies the combined deceleration and s' denotes the distance traveled by the two cars.
Let's assume the combined system’s deceleration is notably higher at (- g).
Utilizing the format V^2 = 2 * 9.8 * 12.25.
This results in V = 15.5 m/s.
Using momentum conservation in the X direction (East): m1 * v1 = (m1+ m2) * V * cos 15.
Solving for v1 gives: v1 = (1650+ 975) Kg * 15.5 m/s * cos 15 /1650 Kg = 23.8 m/s.
Hence, v^2 = V1^2 + 176.4 = (23.8)^2 + 176.4.
This results in v = 27.3 m/s = 61.1 mph (indicating that he exceeded the speed limit when applying the brakes).
In the Y-direction momentum conservation shows: m2 * v2 = (m1+m2) * V * sin 15.
Solving for v2 results in: v2 = (m1+m2) * V * sin 15/m2.
We conclude with v2 = 41.7 m/s (93.3 m/hr) indicating he also exceeded the speed limit.
