Answer:
Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration,
d
d
t
v
(
t
)
=
a
(
t
)
,
ddtv(t)=a(t),
we can take the indefinite integral of both sides, finding
∫
d
d
t
v
(
t
)
d
t
=
∫
a
(
t
)
d
t
+
C
1
,
∫ddtv(t)dt=∫a(t)dt+C1,
where C1 is a constant of integration. Since
∫
d
d
t
v
(
t
)
d
t
=
v
(
t
)
∫ddtv(t)dt=v(t), the velocity is given by
v
(
t
)
=
∫
a
(
t
)
d
t
+
C
1
.
v(t)=∫a(t)dt+C1.
Similarly, the time derivative of the position function is the velocity function,
d
d
t
x
(
t
)
=
v
(
t
)
.
ddtx(t)=v(t).
Thus, we can use the same mathematical manipulations we just used and find
x
(
t
)
=
∫
v
(
t
)
d
t
+
C
2
,
x(t)=∫v(t)dt+C2,
where C2 is a second constant of integration.
We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find
v
(
t
)
=
∫
a
d
t
+
C
1
=
a
t
+
C
1
.
v(t)=∫adt+C1=at+C1.
If the initial velocity is v(0) = v0, then
v
0
=
0
+
C
1
.
v0=0+C1.
Then, C1 = v0 and
v
(
t
)
=
v
0
+
a
t
,
v(t)=v0+at,
which is (Equation). Substituting this expression into (Figure) gives
x
(
t
)
=
∫
(
v
0
+
a
t
)
d
t
+
C
2
.
x(t)=∫(v0+at)dt+C2.
Doing the integration, we find
x
(
t
)
=
v
0
t
+
1
2
a
t
2
+
C
2
.
x(t)=v0t+12at2+C2.
If x(0) = x0, we have
x
0
=
0
+
0
+
C
2
;
x0=0+0+C2;
so, C2 = x0. Substituting back into the equation for x(t), we finally have
x
(
t
)
=
x
0
+
v
0
t
+
1
2
a
t
2
,
x(t)=x0+v0t+12at2,
