A collision occurred between a 1540 kg automobile (A) struck a stopped 890 kg automobile (B). All of the wheels of B were locked, and it slid 6 m after the impact. The coefficient of kinetic friction between B's tires and road was estimated to be pk= 0.8 and the coefficient of restitution of the impact to be e - 0.2. Determine automobile A's velocity in meters per second, just before the impact? Note: You can assume that there was a single impact. B

PROBLEM 1: A car travels a 10-degree inclined road at a speed of 20 ft/s. The driver then applies the break and tires skid marks were made on the pavement at a distance "s". If the coefficient of kinetic friction between the wheels of the 3500-pound car and the road is 0.5, determine the skid mark distance.

Vehicle performance is being tested on a large flat paved area. For this situation, what a coefficient of side friction must be developed to hold a car going 90 kph on a radius of 300 m.

The 4.2 Mg truck and 2.0 Mg car are traveling with the free- rolling velocities shown just before they collide. After the collision, the car moves with a velocity of 16 km/h to the right relative to Part A the truck. Determine the coefficient of restitution between the truck and car.(Figure 1) VO AZ 1 vec ? You have already submitted this answer. Enter a new answer. No credit lost. Try again. Submit Previous Answers Request Answer Part B Determine the loss of energy due to the collision. VO AZ I vec ? Figure 1 of 1 ΔΕ - kJ Submit Previous Answers Request Answer 30 km/h X Incorrect; Try Again; 5 attempts remaining 10 km/h Provide Feedback

A 900-kg car travelling east at 15.0 m/s collides with a 750-kg car travelling north at 20.0 m/s The cars stick together. Assume that any other unbalanced forces are negligible What is the change in momentum of the colliding bodies? What is the speed of the wreckage just after the collision? In what direction does the wreckage move just after the collision?

Problem 1. In deriving braking distance, we assume a constant acceleration/deceleration rate, a. In fact, AASHTO assumes constant acceleration a(t): = c (constant) in most of its design procedures. Although this is not entirely true, distances required for acceleration or deceleration can be realistic if appropriate values are chosen for the constant acceleration process. Linear acceleration a(t) = c₁t (t is time elapse and c₁ is jerk rate or rate of change) is a better approximation to the acceleration process. a) What is the practical stopping distance for a vehicle with initial speed 70 mph, if acceleration is described by a linear function of time using a(t) = -1.8t (t is time elapse in seconds). b) Use a constant acceleration rate to achieve the same stopping distance. What is this constant acceleration rate? Problem 2. A car hits a tree at an estimated speed of 25 mph on a 3 % upgrade. If skid marks of 120 ft are observed on dry pavement (coefficient of braking friction, fo…

Problem #4 (. A 16-lb bowling ball is cast horizontally onto a lane such that its initial angular rate wo = 0 and its center of mass has a velocity vo = 8 ft/s. The radius of the ball is r = 0.375 ft, and the coefficient of kinetic friction between the lane and ball is Hk = 0.12. 8 ft/s G What is the condition, in terms of w and v for а. 0.375 ft rolling without slipping? b. ) Draw a clear and complete free body diagram of just the ball. с. What is the angular acceleration of the ball? Use IG = l5 mr for a solid sphere. What is the linear acceleration of the ball? е. For how much time does the ball slide before it begins to roll without slip? f. For what distance does the ball slide before it begins to roll without slip? g. What is the velocity of the ball at the moment it beings to roll without slip? d.