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The driver has a speed
v
0
and is at position
x
=0 when he sees the deer at
x
=35. After 0.5 s
his position is
0
0
5
.
0
v
x
=
. So, now, the equations of motion are
t
v
v
t
t
v
v
x
10
5
5
.
0
0
2
0
0

=

+
=
.
We are told that when
,
0
=
v
35
=
x
. So,
t
v
10
0
=
and therefore
2
2
2
5
5
5
10
5
35
t
t
t
t
t
+
=

+
=
and so

=
2
.
3
2
.
2
t
seconds (solving the quadratic
equation). Now, there are two solutions which will have the car stop at
35
=
x
m,

=
32
22
0
v
m/s. We know that
0
0
<
v
cannot be true, so the positive time solution must
be the right one. The correct equations of motion for the car thus become
t
v
t
t
x
10
22
5
22
11
2

=

+
=
.
For these equations of motion, there is only one time when
0
=
v
because a parabola has only one
extremum. The graph shows the car
starting at
11
=
x
m and stopping at
35
=
x
m at
2
.
2
=
t
s.
0
0.5
1
1.5
2
2.5
t
0
10
20
30
40
x
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View Full DocumentJust for the fun of it, let’s examine the other solution,
2
.
3

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 Fall '09
 geller

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