2188 Please show all work and BoxCircle all Answers Train A

2/188) Please show all work and Box/Circle all Answers!

Train A travels with a constant speed V_A = 90 km/h along the straight and level track. The driver of car B, anticipating the railway grade crossing C, decreases the car speed of 80 km/h at the rate of 2.6 m/s^3. Determine the velocity v_A/B and acceleration a_A/B of the train relative to the car.

Solution

Speed of train v = 90 km/h

Velocity of train V = v cos 14 i + v sin 14 j

                          = 90 cos 14 i + 90 sin 14 j

                          = 87.32 i + 21.77 j

Speed of car v \' = 80 km/h

Velocity of car V \' = v \' cos 67 i + v \' sin 67 j

                          = 80 cos 67 i + 80 sin 67 j

                          = 31.25 i + 73.64 j

Accleration of train a = 0 i + 0 j    Since it moves with constant speed

Accleration of car a \' = -2.6 cos 67 i - 2.6 sin 67 j

                              = -1.0159 i - 2.3933 j

Velocity of train relative to car = V - V \'

                                           = 87.32 i + 21.77 j -(31.25 i + 73.64 j )

                                          = 56.07 i -51.87 j

Accleration of train relative to car = a - a \'

                                                = ( 0 i + 0 j) -( -1.0159 i - 2.3933 j )

                                                = +1.0159 i + 2.3933 j