Did you draw a diagram(label what happens where and when , velocityInital= ? , velocityFinal=? , Acceleration=? StartPosition =? , EndPosition=? That always helps me in addition to putting closely related equations next to the drawing.
Beh just figured out what I did those times are when the train hits 150m. NOT what the question asked.
Ask for When the ends of the trains are meet.
They will meet when xA+xB=300m.
Consider the starting point at when trains front ends face one another. some velocity on each train side.
300 =...
Hm.. If I'm understanding this correctly:
Consider front of A the origin. Back of A is -150m from origin.
front of B is the origin. Back of B is +150m from origin.
The time in which I solve tells me the following:
Okay, now let's find time that train A went 150m
x =...
Relook agained and though about displacement.
So what I did for solving A)
A) How far does train on the left travel before the front ends of the trains pass?
xA + xB = 40m
x = 1/2*a*t^2
Let's solve for time..5*(1)(t^2) +.5*1.3*t^2= 40
t = 5.89s
Plug back into xA = .5*1*5.89 = 17.34m <-...
Oh, I tried this as well.
xA = 1/2*a*t^2
xB = 1/2*a*t^2
Solve for the time when their positions are equal.
.5*1*t^2 = .5*1.3*t^2 . Cannot get t.
So i tried this since I got velocity average for A
xA = v*t , xB = 1/2*a*t^2, solve for t.
4.47(t) = .5*1.3*t^2
4.47/.5*1.3 = t ; t =...
1. Problem Statement
Two trains face each other, adjacent tracks.
Both at rest, and front ends are 40m apart.
Left train accelerates at 1.0m/s^2.
Right train accelerates at 1.3m/s^2.
A) How far does train on the left travel before the front ends of the trains pass?
B) If trains are each 150m...