385
Answers
601–700
Answers and Explanations
−
+ −
+
+ −
= − − +
∫
∫
t
t
dt
t
t
dt
t
t
t
2
2
2
8
0
2
3
2
0
2
2 6
2
2 6
6
6
2 2
3
2
2
8
3
2
3
2
6
6
2
6
2 6 2
0
8
6
8 6
+
+ −
= − − +
−
+
+ −
t
t
t
( ) ( )
( ( )
( )
8
2
6
2 6 2
344
3
3
2
−
+ −
=
634.
235
6
Begin by finding the velocity function by evaluating the antiderivative of the
acceleration function, a
(t) = 2t + 1:
v t
t
dt
t t C
( )
(
)
=
+
= + +
∫ 2 1
2
Next, use the initial condition, v
(0) = –12, to solve for the arbitrary constant of
integration:
v
C
C
C
( )
0 0 0
12 0 0
12
2
2
= + +
− = + +
− =
Therefore, the velocity function is
v t
t t
( ) = + −
2
12
To find the distance traveled, integrate the absolute value of the velocity function over
the given interval:
t t
dt
2
0
5
12
+ −
∫
Find any zeros of the function on the given interval so you can determine where the
velocity function is positive or negative: t
2
+ t – 12 = 0 factors as (t + 4)(t – 3) = 0 and has
the solutions t = –4, t = 3. The velocity function is negative on the interval (0, 3) and
positive on the interval (3, 5), so the distance traveled is
−
+ −
(
) +
+ −
(
)
= − − +
+
+
∫
∫
t t
dt
t t
dt
t
t
t
t
t
2
0
3
2
3
5
3
2
0
3
3
2
12
12
3 2
12
3 2 2
12
3
3
3
2
12 3
0
5
3
5
2
12 5
3
5
3
2
3
2
−
= −
−
+
−
+
+
−
t
( ) ( )
( ) −
+
−
=
3
3
3
2
12 3
235
6
3
2
( )
Answers
601–700
Answers and Explanations
−
+ −
+
+ −
= − − +
∫
∫
t
t
dt
t
t
dt
t
t
t
2
2
2
8
0
2
3
2
0
2
2 6
2
2 6
6
6
2 2
3
2
2
8
3
2
3
2
6
6
2
6
2 6 2
0
8
6
8 6
+
+ −
= − − +
−
+
+ −
t
t
t
( ) ( )
( ( )
( )
8
2
6
2 6 2
344
3
3
2
−
+ −
=
634.
235
6
Begin by finding the velocity function by evaluating the antiderivative of the
acceleration function, a
(t) = 2t + 1:
v t
t
dt
t t C
( )
(
)
=
+
= + +
∫ 2 1
2
Next, use the initial condition, v
(0) = –12, to solve for the arbitrary constant of
integration:
v
C
C
C
( )
0 0 0
12 0 0
12
2
2
= + +
− = + +
− =
Therefore, the velocity function is
v t
t t
( ) = + −
2
12
To find the distance traveled, integrate the absolute value of the velocity function over
the given interval:
t t
dt
2
0
5
12
+ −
∫
Find any zeros of the function on the given interval so you can determine where the
velocity function is positive or negative: t
2
+ t – 12 = 0 factors as (t + 4)(t – 3) = 0 and has
the solutions t = –4, t = 3. The velocity function is negative on the interval (0, 3) and
positive on the interval (3, 5), so the distance traveled is
−
+ −
(
) +
+ −
(
)
= − − +
+
+
∫
∫
t t
dt
t t
dt
t
t
t
t
t
2
0
3
2
3
5
3
2
0
3
3
2
12
12
3 2
12
3 2 2
12
3
3
3
2
12 3
0
5
3
5
2
12 5
3
5
3
2
3
2
−
= −
−
+
−
+
+
−
t
( ) ( )
( ) −
+
−
=
3
3
3
2
12 3
235
6
3
2
( )