379
Answers
601–700
Answers and Explanations
624.
−40
3
To find the displacement, simply integrate the velocity function over the given interval:
s
s
t
d t
t
t
( ) ( )
()
( )
25
1
4
2
3
4
2
3
25
4 25
1 2
1
25
3 2
1
25
3
−
=
−
(
) =
−
=
( ) −
∫
− −
−
( )
=
−
− +
=
−
= −
2
3
4
250
3
100 2
3
4
248
3
96
40
3
625.
13
Note that displacement, or change in position, can be positive or negative or zero. You
can think of the particle moving to the left if the displacement is negative and moving
to the right if the displacement is positive.
Velocity is the rate of change in displacement with respect to time, so if you integrate
the velocity function over an interval where the velocity is negative, you’re finding how
far the particle travels to the left over that time interval (the value of the integral is negative to indicate that the displacement is to the left). Likewise, if you integrate the velocity function over an interval where the velocity is positive, you’re finding the distance
that the particle travels to the right over that time interval (in this case, the value of the
integral is positive). By combining these two values, you find the net displacement.
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.
v t dt
( )
0
5
∫
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 2t – 4 = 0 so that t = 2.
Because the velocity function is negative on the interval (0, 2) and positive on the
interval (2, 5), the distance traveled is
− −
+
−
= − +
(
) + −
(
)
= − ( ) + −
∫
∫
(
)
(
)
( ) (
2 4
2 4
4
4
2
4 2
0
2
2
5
2
0
2
2
2
5
2
t
dt
t
dt
t
t
t
t
0 0 0
5 4 5
2 4 2
4 9
13
2
2
+
+
−
(
) − −
(
)
= +
=
)
( )
( )
Answers
601–700
Answers and Explanations
624.
−40
3
To find the displacement, simply integrate the velocity function over the given interval:
s
s
t
d t
t
t
( ) ( )
()
( )
25
1
4
2
3
4
2
3
25
4 25
1 2
1
25
3 2
1
25
3
−
=
−
(
) =
−
=
( ) −
∫
− −
−
( )
=
−
− +
=
−
= −
2
3
4
250
3
100 2
3
4
248
3
96
40
3
625.
13
Note that displacement, or change in position, can be positive or negative or zero. You
can think of the particle moving to the left if the displacement is negative and moving
to the right if the displacement is positive.
Velocity is the rate of change in displacement with respect to time, so if you integrate
the velocity function over an interval where the velocity is negative, you’re finding how
far the particle travels to the left over that time interval (the value of the integral is negative to indicate that the displacement is to the left). Likewise, if you integrate the velocity function over an interval where the velocity is positive, you’re finding the distance
that the particle travels to the right over that time interval (in this case, the value of the
integral is positive). By combining these two values, you find the net displacement.
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.
v t dt
( )
0
5
∫
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 2t – 4 = 0 so that t = 2.
Because the velocity function is negative on the interval (0, 2) and positive on the
interval (2, 5), the distance traveled is
− −
+
−
= − +
(
) + −
(
)
= − ( ) + −
∫
∫
(
)
(
)
( ) (
2 4
2 4
4
4
2
4 2
0
2
2
5
2
0
2
2
2
5
2
t
dt
t
dt
t
t
t
t
0 0 0
5 4 5
2 4 2
4 9
13
2
2
+
+
−
(
) − −
(
)
= +
=
)
( )
( )