A particle starts with an initial velocity u and gained a final velocity v
with uniform acceleration a. In due course it covered a displacement
S. Write an equation relating u,v,a, S.
Answers
Answer: This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration, we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position function we found the velocity function, and likewise, by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function.
Kinematic Equations from Integral Calculus
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
)
,
we can take the indefinite integral of both sides, finding
∫
d
d
t
v
(
t
)
d
t
=
∫
a
(
t
)
d
t
+
C
1
,
where C1 is a constant of integration. Since
∫
d
d
t
v
(
t
)
d
t
=
v
(
t
)
, the velocity is given by
v
(
t
)
=
∫
a
(
t
)
d
t
+
C
1
.
Similarly, the time derivative of the position function is the velocity function,
d
d
t
x
(
t
)
=
v
(
t
)
.
Thus, we can use the same mathematical manipulations we just used and find
x
(
t
)
=
∫
v
(
t
)
d
t
+
C
2
,
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
.
If the initial velocity is v(0) = v0, then
v
0
=
0
+
C
1
.
Then, C1 = v0 and
v
(
t
)
=
v
0
+
a
t
,
which is (Equation). Substituting this expression into (Figure) gives
x
(
t
)
=
∫
(
v
0
+
a
t
)
d
t
+
C
2
.
Doing the integration, we find
x
(
t
)
=
v
0
t
+
1
2
a
t
2
+
C
2
.
If x(0) = x0, we have
x
0
=
0
+
0
+
C
2
;
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
,
which is (Equation).