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0.5
1
1.5
2
2.5
20
40
60
80
100
120
t
= 1, as the automobile cannot suddenly jump from one position to a completely diFerent position in an
instant. Hence 45 = 75 +
C
, so that
C
=
−
30. Therefore
x
(
t
)=
½
45
t
if 0
5
t
5
1;
75
t
−
30
if 1
<t
5
2
.
To see the graph of
x
(
t
), plot in
Mathematica
x[t
]:=I
f[t
<
1,45
∗
t,75
∗
t
−
30 ]
on the interval 0
5
t
5
2.
C01S02.074:
The graph of
x
(
t
) will consist of three straightline segments (because of the constant speeds),
the ±rst of slope 60 for 0
5
t
5
1, the second of slope zero for 1
5
t
5
1
.
5, and the third of slope 60 for
1
.
5
5
t
5
2
.
5. The ±rst pair must coincide when
t
= 1 and the second pair must coincide when
t
=1
.
5
because the graph of
x
(
t
) can have no discontinuities. So if we write
x
(
t
) = 60 for 0
5
t
5
1, we must have
x
(
t
) = 60 for 1
5
t
5
1
.
5. ²inally,
x
(
t
)=60
t
+
C
for some constant
C
if 1
.
5
5
t
5
2
.
5, but the latter must
equal 60 when
t
=1
.
5, so that
C
=
−
30. Hence
x
(
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This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.
 Spring '11
 CHENG
 Calculus

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