18. अंत मे परसाई जी सीधा स्टेशन जाने का निश्चय क्यों लिया
answer I want
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Explanation:
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,
\[\frac{d}{dt}v(t)=a(t),\]
we can take the indefinite integral of both sides, finding
\[\int \frac{d}{dt}v(t)dt=\int a(t)dt+{C}_{1},\]
where C1 is a constant of integration. Since
\[\int \frac{d}{dt}v(t)dt=v(t)\]
, the velocity is given by
\[v(t)=\int a(t)dt+{C}_{1}.\]
Similarly, the time derivative of the position function is the velocity function,
\[\frac{d}{dt}x(t)=v(t).\]
Thus, we can use the same mathematical manipulations we just used and find
\[x(t)=\int v(t)dt+{C}_{2},\]
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
\[v(t)=\int adt+{C}_{1}=at+{C}_{1}.\]
If the initial velocity is v(0) = v0, then
\[{v}_{0}=0+{C}_{1}.\]
Then, C1 = v0 and
\[v(t)={v}_{0}+at,\]
which is (Equation). Substituting this expression into (Figure) gives
\[x(t)=\int ({v}_{0}+at)dt+{C}_{2}.\]
Doing the integration, we find
\[x(t)={v}_{0}t+\frac{1}{2}a{t}^{2}+{C}_{2}.\]
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
\[x(t)={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2},\]
which is (Equation).