Question
Obtain equations of motion for constant acceleration using method of calculus.
✓
Answer
By definition
$
\begin{array}{l}
a=\frac{ d v}{ d t} \\
d v=a d t
\end{array}
$
Integrating both sides
$
\begin{aligned}
\int_{t_0}^v d v & =\int_0^t a d t \\
& =a \int_0^t d t \quad \text{(a is constant)}
\end{aligned}
$
$
\begin{aligned}
v-v_0 & =a t \\
v & =v_0+a t
\end{aligned}
$
Further,
$
\begin{aligned}
v & =\frac{ d x}{ d t} \\
d x & =v d t
\end{aligned}
$
Integrating both sides
$
\int_{x_0}^x d x=\int_0^t v d t
$
$
\begin{aligned}
& =\int_0^t\left(v_0+a t\right) d t \\
x-x_0 & =v_0 t+\frac{1}{2} a t^2 \\
x & =x_0+v_0 t+\frac{1}{2} a t^2
\end{aligned}
$
We can write
$
a=\frac{ d v}{ d t}=\frac{ d v}{ d x} \frac{ d x}{ d t}=v \frac{ d v}{ d x}
$
or, $v d v=a d x$
Integrating both sides,
$
\begin{array}{l}
\int_{ L _0}^v v d v=\int_{x_0}^x a d x \\
\frac{v^2-v_0^2}{2}=a\left(x-x_0\right) \\
v^2=v_0^2+2 a\left(x-x_0\right)
\end{array}
$
$
\begin{array}{l}
a=\frac{ d v}{ d t} \\
d v=a d t
\end{array}
$
Integrating both sides
$
\begin{aligned}
\int_{t_0}^v d v & =\int_0^t a d t \\
& =a \int_0^t d t \quad \text{(a is constant)}
\end{aligned}
$
$
\begin{aligned}
v-v_0 & =a t \\
v & =v_0+a t
\end{aligned}
$
Further,
$
\begin{aligned}
v & =\frac{ d x}{ d t} \\
d x & =v d t
\end{aligned}
$
Integrating both sides
$
\int_{x_0}^x d x=\int_0^t v d t
$
$
\begin{aligned}
& =\int_0^t\left(v_0+a t\right) d t \\
x-x_0 & =v_0 t+\frac{1}{2} a t^2 \\
x & =x_0+v_0 t+\frac{1}{2} a t^2
\end{aligned}
$
We can write
$
a=\frac{ d v}{ d t}=\frac{ d v}{ d x} \frac{ d x}{ d t}=v \frac{ d v}{ d x}
$
or, $v d v=a d x$
Integrating both sides,
$
\begin{array}{l}
\int_{ L _0}^v v d v=\int_{x_0}^x a d x \\
\frac{v^2-v_0^2}{2}=a\left(x-x_0\right) \\
v^2=v_0^2+2 a\left(x-x_0\right)
\end{array}
$
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