Part II: The Answers
380
Answers
601–700
626.
5
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 t
dt
2
2
4
6
− −
∫
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 t
2
– t – 6 = 0, or (t – 3)
(t + 2) = 0, which has solutions t = 3 and t = –2. Because the velocity function is negative
on the interval (2, 3) and positive on the interval (3, 4), the distance traveled is
−
− −
(
) +
− −
(
)
= − + +
+
− −
∫
∫ t t dt
t t
dt
t
t
t
t
t
2
2
3
4
2
3
3
2
2
3
3
2
6
6
3 2
6
3 2
6t t
= −
+
+
− −
+
+
+
−
3
4
3
2
3
2
3
2
3
3
3
2
6 3
2
3
2
2
6 2
4
3
4
2
( )
( )
− −
−
−
−
=
6 4
3
3
3
2
6 3
5
3
2
( )
( )
627.
4
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.
2
0
cos t dt
π
∫
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 2 cos t ≥ 0 on 0 2
, π
( )
and 2 cos t ≤ 0 on π π
2
,
( ) . Therefore, the distance traveled is
2
2
2
2
2
2
2
0
2
0
2
2
0
2
2
cos
c os
sin
s in
sin
sin
s
t
t dt
t
t
π
π
π
π
π
π
π
∫
∫
+
−
=
−
=
−
(
) − i in sin
π
π
−
( )
= +
=
2
2
2 2
4
380
Answers
601–700
626.
5
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 t
dt
2
2
4
6
− −
∫
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 t
2
– t – 6 = 0, or (t – 3)
(t + 2) = 0, which has solutions t = 3 and t = –2. Because the velocity function is negative
on the interval (2, 3) and positive on the interval (3, 4), the distance traveled is
−
− −
(
) +
− −
(
)
= − + +
+
− −
∫
∫ t t dt
t t
dt
t
t
t
t
t
2
2
3
4
2
3
3
2
2
3
3
2
6
6
3 2
6
3 2
6t t
= −
+
+
− −
+
+
+
−
3
4
3
2
3
2
3
2
3
3
3
2
6 3
2
3
2
2
6 2
4
3
4
2
( )
( )
− −
−
−
−
=
6 4
3
3
3
2
6 3
5
3
2
( )
( )
627.
4
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.
2
0
cos t dt
π
∫
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 2 cos t ≥ 0 on 0 2
, π
( )
and 2 cos t ≤ 0 on π π
2
,
( ) . Therefore, the distance traveled is
2
2
2
2
2
2
2
0
2
0
2
2
0
2
2
cos
c os
sin
s in
sin
sin
s
t
t dt
t
t
π
π
π
π
π
π
π
∫
∫
+
−
=
−
=
−
(
) − i in sin
π
π
−
( )
= +
=
2
2
2 2
4