A body has displacement x= √2+t^2m. find velocity at time 2 second
Answers
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
)
,
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
,