381
Answers
601–700
Answers and Explanations
628.
4 2
3 1
2
−
−
Unlike displacement, distance can’t be negative. In order to find the distance traveled,
you can integrate the absolute value of the velocity function because you’re then
integrating a function that’s greater than or equal to zero on the given interval.
sin cos
t
t dt
−
−
∫ π
π
6
2
Find any zeros of the function on the given interval so you can determine where the
velocity function is positive or negative. In this case, you have sin t – cos t = 0 so that
sin t = cos t, which has a solution t = π
4
on the given interval. Note that on the interval
−
( )
π π
6 4
,
, you have sin t – cos t < 0 and that on the interval π π
4 2
,
( ) , sin t – cos t > 0.
Therefore, the distance traveled is
−
−
+
−
=
+
+
−
−
∫
∫
(sin cos )
(sin cos )
(cos sin )
(
t
t dt
t
t dt
t
t
≠
≠
≠
≠
≠
≠
6
4
4
2
6
4
− −
−
=
+
(
) − − + −
(
) −
+
cos sin )
cos
sin
cos
sin
cos
sin
t
t ≠
≠
≠
≠
≠
≠
≠
≠
4
2
4
4
6
6
2
2
( (
) − +
(
)
=
+
−
+ −
+ −
+
cos
sin
(
)
≠
≠
4
4
2
2
2
2
3
2
1
2
0 1
2
2
2
2
=
+
−
+ − +
+
=
−
−
2
2
2
2
3
2
1
2
1
2
2
2
2
4 2
3 1
2
629.
68
3
Unlike displacement, distance can’t be negative. In order to find the distance traveled,
you can integrate the absolute value of the velocity function because you’re then integrating a function that’s greater than or equal to zero on the given interval.
t
dt
−
∫
4
1
25
Find the any zeros of the function on the given interval so you can determine where
the velocity function is positive or negative. In this case, you have t − =
4 0 so that
t = 4, or t = 16. Notice that on the interval (1, 16), you have t − <
4 0, whereas on the
interval (16, 25), you have t − >
4 0. Therefore, the distance traveled is
−
−
(
) +
−
(
)
= −
+
+
−
(
∫
∫
t
d t
t
dt
t
t
t
t
1 2
1
16
1 2
16
25
3 2
1
16
3 2
4
4
2
3
4
2
3
4 ) )
= −
+
− − +
( )
+
−
−
16
25
3 2
3 2
2
3
16
4 16
2
3
4
2
3
25
4 25
2
3
16
( )
( )
( )
( )
( )
3 3 2
4 16
128
3
64 10
3
250
3
100 128
3
64
68
3
−
(
)
= −
+ −
+
−
−
+
=
( )
Answers
601–700
Answers and Explanations
628.
4 2
3 1
2
−
−
Unlike displacement, distance can’t be negative. In order to find the distance traveled,
you can integrate the absolute value of the velocity function because you’re then
integrating a function that’s greater than or equal to zero on the given interval.
sin cos
t
t dt
−
−
∫ π
π
6
2
Find any zeros of the function on the given interval so you can determine where the
velocity function is positive or negative. In this case, you have sin t – cos t = 0 so that
sin t = cos t, which has a solution t = π
4
on the given interval. Note that on the interval
−
( )
π π
6 4
,
, you have sin t – cos t < 0 and that on the interval π π
4 2
,
( ) , sin t – cos t > 0.
Therefore, the distance traveled is
−
−
+
−
=
+
+
−
−
∫
∫
(sin cos )
(sin cos )
(cos sin )
(
t
t dt
t
t dt
t
t
≠
≠
≠
≠
≠
≠
6
4
4
2
6
4
− −
−
=
+
(
) − − + −
(
) −
+
cos sin )
cos
sin
cos
sin
cos
sin
t
t ≠
≠
≠
≠
≠
≠
≠
≠
4
2
4
4
6
6
2
2
( (
) − +
(
)
=
+
−
+ −
+ −
+
cos
sin
(
)
≠
≠
4
4
2
2
2
2
3
2
1
2
0 1
2
2
2
2
=
+
−
+ − +
+
=
−
−
2
2
2
2
3
2
1
2
1
2
2
2
2
4 2
3 1
2
629.
68
3
Unlike displacement, distance can’t be negative. In order to find the distance traveled,
you can integrate the absolute value of the velocity function because you’re then integrating a function that’s greater than or equal to zero on the given interval.
t
dt
−
∫
4
1
25
Find the any zeros of the function on the given interval so you can determine where
the velocity function is positive or negative. In this case, you have t − =
4 0 so that
t = 4, or t = 16. Notice that on the interval (1, 16), you have t − <
4 0, whereas on the
interval (16, 25), you have t − >
4 0. Therefore, the distance traveled is
−
−
(
) +
−
(
)
= −
+
+
−
(
∫
∫
t
d t
t
dt
t
t
t
t
1 2
1
16
1 2
16
25
3 2
1
16
3 2
4
4
2
3
4
2
3
4 ) )
= −
+
− − +
( )
+
−
−
16
25
3 2
3 2
2
3
16
4 16
2
3
4
2
3
25
4 25
2
3
16
( )
( )
( )
( )
( )
3 3 2
4 16
128
3
64 10
3
250
3
100 128
3
64
68
3
−
(
)
= −
+ −
+
−
−
+
=
( )