1. Chapter 2
Equations for Motion with Constant Acceleration
No. Equation missing
(1) atvv o += oxx −
(2) 2
2
1
attvxx oo +=− v
(3) )(222
oo xxavv −+= t
(4) tvvxx oo )(2
1
+=− a
(5) 2
2
1
atvtxx o −=− ov
The first two equations (1) and (2) are the essential equations because the other three
equations can be derived from their combinations.
Let us first derive the first two essential equations:
Equation 1:
If the acceleration is constant then the average acceleration and the
instantaneous acceleration are equal.
t
vv
t
vv
aaa oo
avgin
−
=
−
−
===
0
where is the initial velocity at t = 0ov
rewriting the equation
atvvvvat o +=⇒−= 0 (1)
Equation 2:
To Derive equation (2) we need three equations, one of them is equation (1), and
the other two equations are (6) and (8) as follows:
We know that the average of two values can be obtained by adding them and
dividing the result by two, therefore for a linear velocity we can calculate the
average velocity as:
)(
2
2
1
vv
vv
v o
o
avg +=
+
= (6)
For the third equation we can use the standard equation which we learned
previously:
avgv
t
xx
v o
avg
−
= (7)
Now combining both equations (1) and (6) by substituting v in equation (6)
))((2
1
atvvv ooavg ++= (8)
Then replacing vavg by its value from equation (7) into equation (8)
atvv
t
xx
oo
o
2
1
2
1
2
1
++=
−
Multiplying both sides by t then
2
2
1
attvxx oo +=− (2)
Physics 101 10/10/2004 Dr. Abdulnasser Burezq
PhysicsDepartment
2. Equation 3:
Starting from equation (1) : atvv o += and then rewriting it again as :
a
vv
t o−
=
substitute this into equation (2) :
2
2
1
⎟
⎠
⎞
⎜
⎝
⎛ −
+⎟
⎠
⎞
⎜
⎝
⎛ −
=−
a
vv
a
a
vv
vxx oo
oo
2
22
)(
2
1
a
vv
a
a
vvv
xx ooo
o
−
+
−
=−
a
vv
a
vvv
xx ooo
o
2
)( 22
−
+
−
=−
a
vvvv
a
vvv
xx oooo
o
2
)2( 222
+−
+
−
=−
a
vvvvvvv
xx oooo
o
2
222 222
+−+−
=−
a
vv
xx o
o
2
22
−
=−
22
)(2 oo vvxxa −=−
)(222
oo xxavv −+= (3)
Equation 4:
Starting from equation (1) : atvv o += then rewriting it again as :
t
vv
a o−
=
substitute this into equation (2) :
2
2
1
t
t
vv
tvxx o
oo ⎟
⎠
⎞
⎜
⎝
⎛ −
+=−
( )tvvtvxx ooo −+=− 2
1
tvtvtvxx ooo 2
1
2
1
−+=−
tvvxx oo )(2
1
+=− (4)
Equation 5:
Starting from equation (1) : atvv o += then rewriting it again as : atvvo −=
substitute this into equation (2) :
2
2
1
)( attatvxx o +−=−
2
2
12
atatvtxx o +−=−
2
2
1
atvtxx o −=− (5)
Physics 101 10/10/2004 Dr. Abdulnasser Burezq
PhysicsDepartment
