D1 = Distance traveled by train on left.
D2 = Distance traveled by train on right
a. D1 + D2 = 30 m.
0.5a1*t^2 + 0.5a2*t^2 = 30
0.5*1.17*t^2 + 0.5*1.1*t^2 = 30
0.585t^2 + 0.55t^2 = 30.
1.135t^2 = 30.
t^2 = 26.43
t = 5.14s.
D1 = 0.5*a1*t^2.
a1 = 1.17 m/s^2.
t = 5.14 s.
Solve for D1.
b. D1 - D2 = 300 m.
0.5*1.17*t^2 + 0.5*1.1*t^2 = 2*150 = 300
0.585t^2 + 0.55t^2 = 300
1.135t^2 = 300
t^2 = 264.3
t = 16.3 s.
Two trains face each other on adjacent tracks. They are initially at rest, and their front ends are 30 m apart. The train on the left accelerates rightward at 1.17 m/s^2. The train on the right accelerates leftward at 1.1 m/s^2.
(a) How far does the train on the left travel before the front ends of the trains pass?
(b) If the trains are each 150 m in length, how long after the start are they completely past one another, assuming their accelerations are constant?
Please use kinematics equations to solve.
2 answers
Correction:
b. D1 + D2 = 30+300 = 330 m.
0.5*1.17*t^2 + 0.5*1.1*t^2 = 330.
0.585t^2 + 0.55t^2 = 330.
1.135t^2 = 330.
t^2 = 290.75.
t = 17.1 s.
b. D1 + D2 = 30+300 = 330 m.
0.5*1.17*t^2 + 0.5*1.1*t^2 = 330.
0.585t^2 + 0.55t^2 = 330.
1.135t^2 = 330.
t^2 = 290.75.
t = 17.1 s.