This preview shows page 1. Sign up to view the full content.
Unformatted text preview: takes: t=
t= 4v2 ∆x
2
v1f
4 (45mph) (3.5miles)
2 (60mph)
t = 0.175hrs
t = 10.5min This seems to be a reasonable time. Our velocity when we pass the freight train: v = at
2
v1f 4v2 ∆x
2
2∆x v1f
2v2
v=
1
v = 2v2 v= v = 2 (45mph)
v = 90mph
While 90 mph seems high, it is reasonable considering the train is assumed to be undergoing constant acceleration. to solve for acceleration: a1 = 2
2
vf −vi
2∆x It is important to remember that vf is the velocity of our train at the end of our ∆x, 3.5 miles, which is not the
point where the pass. Substituting this into our time equation: t= 2v2 2∆x
2
2
1 vf −vi or, knowing that t= , vi is zero for our train and clearing up the ambiguity of using vf : 4vi ∆x
2
v1f To fi...
View
Full
Document
This document was uploaded on 02/13/2014.
 Spring '14

Click to edit the document details