Example 1

What is the position of an object at t  3 s if its velocity is given by v  2t 2  4 ?
Remember that given a position function we can find the velocity by
evaluating the derivative of the position function with respect to time at the
particular time.

dx


Given x t , then v t  
dt
Hence, what we need is an antiderivative of the velocity function.
 
x   v dt

x   2t 2  4 dt
 2 3
x  t  4t
3


Example 1

What is the position of an object at t  3 s if its velocity is given by v  2t 2  4 ?
 2 3
x  t  4t
3
To check our answer we need only take the derivative of the result.

 dx
v
dt
 d 2 3

v   t  4t 
dt  3


v  2t 2  4
It’s appears we have found the right function …
or have we????
Example 1

What is the position of an object at t  3 s if its velocity is given by v  2t 2  4 ?
 2 3
x  t  4t
3
What if the position function is
 2 3
x  t  4t  5
3
 d 2 3

v   t  4t  5 
dt  3


v  2t 2  4
So this function also works!
Example 1

What is the position of an object at t  3 s if its velocity is given by v  2t 2  4 ?
 2 3
x  t  4t
3
What if the position function is
 2 3
x  t  4t  10
3
 d 2 3

v   t  4t  10 
dt  3


v  2t 2  4
So this function works as well!
Example 1

What is the position of an object at t  3 s if its velocity is given by v  2t 2  4 ?
Since the derivative of a constant is 0, it seems that any function of the form
 2 3
x  t  4t  C
3
where C  constant
works just as well!
Indefinite Integral 
 f  x dx  F  x   C where C is any constant
Example 1

What is the position of an object at t  3 s if its velocity is given by v  2t 2  4 ?
 2 3
x  t  4t  C
3
In order to find a particular solution you must be given an initial condition.

Suppose x 0 s   5 m
2 3

x (0)  5 m  0  40  C
3
C 5m
 2 3
x  t  4t  5 Particular Solution
3
2 3

x 3 s   3  43  5
3

x 3 s   35 m
Example 2
What is the velocity of an object at t  2 s whose acceleration is given by


a   6 m 3 t   6 m 4 t 2 if v 4 s   90. m ?
s

s  
s 


v   a dt



v   6t  6t 2 dt

v  3t 2  2t 3  C
But

2
3
m
v 4 s   90.
 34  24  C
s
C  10. m
s

v  3t 2  2t 3  10.
Example 2
What is the velocity of an object at t  2 s whose acceleration is given by


a   6 m 3 t   6 m 4 t 2 if v 4 s   90. m ?
s

s  
s 

v  3t 2  2t 3  10.

v 2  322  223  10.

v 2  14 m
s
Example 3


The velocity of a particle is given by v t   2.0t 2  3.0t  2.0 and x 0  0 m.
a. What is the particle' s position at t  3.0 s?
 
x   v dt

x   2t 2  3t  2 dt


 2 3 3 2
x  t  t  2t  C
3
2
But
2 3 3 2

x 0  0  0  0  20  C
3
2
C 0
 2 3 3 2
x  t  t  2t
3
2
Example 3


The velocity of a particle is given by v t   2.0t 2  3.0t  2.0 and x 0  0 m.
a. What is the particle' s position at t  3.0 s?
 2 3 3 2
x  t  t  2t
3
2
2
3

3
2
x 3.0  3.0  3.0  23.0
3
2

x 3.0   1.5 m
Example 3


The velocity of a particle is given by v t   2.0t 2  3.0t  2.0 and x 0  0 m.
b. What is the particle' s accelerati on at t  3.0 s?

 dv
a
dt
 d 2
a
2t  3t  2
dt

a  4t  3


a 3.0  43.0  3

a 3.0   9.0 m
s2

Example 3


The velocity of a particle is given by v t   2.0t 2  3.0t  2.0 and x 0  0 m.
c. What is the particle' s displacement from t  2.0 s to t  4.0 s?
 4
x  2 v dt


 4 2
x  2 2t  3t  2 dt
4
 2 3 3 2

x   t  t  2t 
2
3
2
 2 3 3 2
 2 3 3 2

x   4  4  24   2  2  22
2
2
3
 3

2  2

x  10    4 
3  3

x  15 m
Example 4


A ball is thrown upward with speed vo at position xo . What is its position at t  t1?
 
a  g  constant near the surface of the Earth


 a dt   g dt
 
v  gt  C
But



v 0   vo  g 0   C

C  vo
  
v  gt  vo
1st Kinematic Equation
Example 4


A ball is thrown upward with speed vo at position xo . What is its position at t  t1?
  
v  vo  gt 1st Kinematic Equation

 
 v dt   vo  gt  dt
12
 
x  vot  gt  C
2
But
1 2

 
x 0   xo  vo 0   g 0  C
 2
C  xo
12 
 
x  vot  gt  xo
2
Example 4


A ball is thrown upward with speed vo at position xo . What is its position at t  t1?
12 
 
x  vot  gt  xo
2
12
  
x  xo  vot  gt
2
3rd Kinematic Equation
1 2

 
x t1   xo  vot1  gt1
2