Position, Velocity and
Acceleration
3.2
Position, Velocity and
Acceleration

All fall under rectilinear motion




Motion along a straight line
We are normally given a function relating the
position of a moving object with respect to
time.
Velocity is the derivative of position
Acceleration is the derivative of velocity
Position, Velocity and
Acceleration

Position


Velocity


v(t) or s’(t)
Acceleration


s(t) or x(t)
a(t) or v’(t) or s’’(t)
Speed is the absolute value of velocity
Example

If the position of a particle at time t is given by
the equation below, find the velocity and
acceleration of the particle at time, t = 5.
s(t )  t 3 11t 2  24t
v(t )  s' (t )  3t 2  22t  24
v(5)  3(5)  22(5)  24  11
2
a(t )  v' (t )  s' ' (t )  6t  22
a(5)  6(5)  22  8
Position, Velocity and
Acceleration





When velocity is negative, the particle is moving to
the left or backwards
When velocity is positive, the particle is moving to
the right or forwards
When velocity and acceleration have the same sign,
the speed is increasing
When velocity and acceleration have opposite signs,
the speed is decreasing.
When velocity = 0 and acceleration does not, the
particle is momentarily stopped and changing
direction.
Example

If the position of a particle is given below, find
the point at which the particle changes
direction.
s(t )  t 12t  36t  18
3
2
v(t )  3t 2  24t  36
a(t )  6t  24
a(6)  12
a(2)  12
Changes direction
when velocity = 0
and acceleration
does not
0  3t 2  24t  36
0  t 2  8t  12
0  (t  6)(t  2)
t  6, 2
Example

Using the previous function, find the interval
of time during which the particle is slowing
down.
V(t) = 0 at 2 and
6, a(t) = 0 at 4
s(t )  t 12t  36t  18
3
2
v(t )  3t 2  24t  36
a(t )  6t  24
0
Particle is slowing
down when,
0<t<2
4<t<6
t
v(t)
a(t)
4
2
+
-
-
6
+
+
+
Example
When velocity = 0
When does this occur?

How far does a particle travel between the
eighth and tenth seconds if its position is
given by: s(t )  t 2  6t
To find the total
distance we
must find if the
particle
changes
directions at
any time in the
interval
v(t )  2t  6
0  2t  6
3t
The object may travel forward
then backwards, thus s(10) –
s(8) is really only the
3 displacement
is not in our interval
it will
not theso
total
not affect
our problem!
distance!
s(10)  s(8)  40 16  24
Example

How far does a particle travel between zero
and four seconds if its position is given by:
s(t )  t 4  8t 2
v(t )  4t 3 16t
s(2)  s(0)  s(4)  s(2)
 16  32  12816
v(t )  4t (t 2  4)
0  4t (t  2)(t  2)
 160
t  0, 2,  2
Divide into intervals; 02 and 24
At any time t, the position of a particle moving along an
axis is: s(t )  t 3  6t 2  9t
A. Find the body’s acceleration each time the velocity is zero
v(t )  3t 12t  9
s(1) 0s(0t)2  4s(t2) 3 s(1)
2
C. Find the total distance
traveled by the body from t = 0 to t = 2
 4 00 (2t  34)(t  1)
 6 t  3, 1
a(t )  6t  12
a(1Velocity
)  6= 0 at 1!
a(3)  6
Divide into intervals; 01 and 12
B. Find the body’s speed each time the acceleration is zero
a(t )  6t  12
0  6t  12
t2
v(2)  12  24  9  3
At any time t, the position of a particle moving along an
3
axis is:
t
2
s(t ) 
3
 2t  3t
A. When is the body moving forward? backwards?
v(t )  t 2  4t  3
v(t)
0  (t  3)(t  1)
t  3, 1
1
3
+
-
+
Forward (0, 1) and (3, ∞)
Backwards from (1, 3)
B. When is the velocity increasing? decreasing?
a(t )  2t  4
Velocity increasing:
(1, 2) and (3,
t ∞)2
Velocity decreasing:
(0, 1) and (2, 3)
0
t
v(t)
a(t)
1
+
-
2
-
3
+
+
+