# Derivative Examples 2 ```Derivative Examples 2

x (m)
10
Example 3
8
6
Complete the velocity vs. time and acceleration vs.
time graphs for an object whose motion results in the
position vs. time graph below.
4
2
0
2
4
6
8
2
4
6
8
2
4
6
8
t (s )
-2
-4


v  slope of x vs. t graph

 x
v
t
4s t  6 s
 8m
v
2s

v  4m
s

v (m )
s
4
2
0
t (s )
-2
-4

a (m
s2
)
4
0
-4
t (s )

x (m)
Derivative Examples 2
10
Example 3
8
6
Complete the velocity vs. time and acceleration vs.
time graphs for an object whose motion results in the
position vs. time graph below.
4
2
0
2
4
6
8
2
4
6
8
2
4
6
8
t (s )
-2
-4


a  slope of v vs. t graph



v

v


a
a
t
t

a
4m
s
2s

a  2m
s2

a

v (m )
s
4
2
0
4m
-2
s
1s

a  4 m
t (s )
-4

a (m
s2
)
4
s2
0
-4
t (s )

x (m)
Derivative Examples 2
10
Example 3
8
6
Complete the velocity vs. time and acceleration vs.
time graphs for an object whose motion results in the
position vs. time graph below.
4
2
0
4
6
8
2
4
6
8
2
4
6
8
t (s )
-2


v  slope of x vs. t graph
-4

v (m )
s

 dx
v
dt
4
2
0


a  slope of v vs. t graph

 dv
a
dt
2
 d x
a 2
dt
2
t (s )
-2
-4

a (m
s2
)
4
The second derivative of the
position with respect to time
0
-4
t (s )
Derivative Examples 2

x (m)
10
Example 3
8
6
Complete the velocity vs. time and acceleration vs.
time graphs for an object whose motion results in the
position vs. time graph below.


v  slope of x vs. t graph
 x
v
 The slope of a line connecting
t

any two points on a x vs. t graph.
4
2
0
2
4
6
8
2
4
6
8
2
4
6
8
t (s )
-2
-4

v (m )
s
4
2
0
t (s )
-2


a  slope of v vs. t graph
 v
a
 The slope of a line connecting
t

any two points on a v vs. t graph.
-4

a (m
s2
)
4
0
-4
t (s )
Derivative Examples 2
Example 4

An object moves with position given by xt   2.0t 3  3.0 calculate the object's
a. Instantane ous velocity at t  1.0 s and t  2.0 s.

 dx
v
dt

 d 2.0t 3  3.0
v
dt



 d 2.0t 3 d 3.0
v

dt
dt

v  6.0t 2


2
2
v 1.0 s   6.01.0
v 2.0 s   6.02.0


m
m
v 1.0 s   6.0


v
2
.
0
s

24
s
s
Derivative Examples 2
Example 4 continued

An object moves with position given by xt   2.0t 3  3.0 calculate the object's
b. Instantane ous accelerati on at t  1.0 s and t  2.0 s.

 dv
a
dt

 d 6.0t 2
a
dt

a  12t

a1.0 s   121.0

a 1.0 s   12 m 2
s


a2.0 s   122.0

a 2.0 s   24 m 2
s
Derivative Examples 2
Example 4 continued

An object moves with position given by xt   2.0t 3  3.0 calculate the object's
c. Average velocity and average accelerati on from t  1.0 s and t  2.0 s.
 x
v
t
 x 2.0 s   x 1.0 s 
v
2.0 s  1.0 s


 2.02.03  3.0  2.01.03  3.0
v
1.0 s

v  14 m
s
Part (a)
Notice: Not the simple average of (a)
since the velocity function is nonlinear
and the rate of change of the velocity is
non-constant.

Derivative Examples 2
Example 4 continued

An object moves with position given by xt   2.0t 3  3.0 calculate the object's
c. Average velocity and average accelerati on from t  1.0 s and t  2.0 s.
 v
a
t
 v 2.0 s   v 1.0 s 
a
2.0 s  1.0 s
 24 m s  6.0 m s
a
1.0 s

a  18 m 2
s
Part (b)
Notice: This is the simple average of (b)
since the acceleration function is linear so
that the rate of change of the acceleration
is constant.
```