This preview shows page 1. Sign up to view the full content.
2.93:
The velocities are
2
3
2
and
2
t
t
v
t
v
B
A
δ
γ
β
α

=
+
=
a) Since
v
B
is zero at
t
= 0, car
A
takes the early lead.
b) The cars are both at the origin at
t
= 0.
The nontrivial solution
is found by setting
x
A
=
x
B
, cancelling the common factor of
t
, and solving the quadratic
for
[
]
.
4
)
(
)
(
2
1
2
αδ


±

=
t
Substitution of numerical values gives 2.27 s, 5.73 s.
The use of the term “starting point”
can be taken to mean that negative times are to be neglected.
This is the end of the preview. Sign up
to
access the rest of the document.