Question
Derive the three equations of motion by calculus method. Express conditions under which they can be used.
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Answer
Consider an object moving in straight line with uniform acceleration $= a$. Let at $t = 0$ velocity of the body $= u$ at $t = t$ velocity of the body $= v$
$\text{dv}=\text{a dt}$
Integrating from $0 \rightarrow t$ when velocity changes from $u \rightarrow v$
$\int\limits^\text{v}_\text{u}\text{dv}=\text{a}\int\limits^{\text{t}}_\text{0}\text{dt}$
$\text{v}-\text{u}=\text{at}$
$\text{v}=\text{u}+\text{at}\ \dots(\text{i})$
$\text{dx}=\text{vdt}$
$\text{dx}=(\text{u}+\text{at})\text{dt} [$from $(i) \text{v}=\text{u}+\text{at}]$
Let $x_0 =$ displacement at $t = 0$
$x =$ displacement at $t = t$
Integrating within limits
$\int\limits^\text{x}_{\text{x}_0}\text{dx}=\int\limits^{\text{t}}_0(\text{u}+\text{at})\text{dt}$
$=\text{u}\int\limits^{\text{t}}_0\text{dt}+\text{a}\int\limits^{\text{t}}\text{tdt}$
$\text{x}-\text{x}_0=\text{ut}+\frac{1}{2}\text{at}^2$
$\text{x}=\text{x}_0+\text{ut}+\frac{1}{2}\text{at}^2\ \dots(\text{ii})$
If $x - x_0 = s$ = distance covered by an object in time $t$ then
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
then $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{dv}}{\text{dx}}\times\frac{\text{dx}}{\text{dt}}=\text{v}\frac{\text{dv}}{\text{dx}}$
$\text{adx}=\text{vdv}$
Let $u$ be the velocity of object at position $x_0\ v$ be the velocity of object at position $x$
Integrating above within limits
$\int\limits^{\text{x}}_{\text{x}_0}\text{a dx} = \int\limits^{\text{c}}_{\text{u}}\text{v dv}$
$\text{a}(\text{x}-\text{x}_0)=\frac{\text{v}^2}{2}-\frac{\text{u}^2}{2}$
$\text{v}^2-\text{u}^2=2\text{a}(\text{x}-\text{x}_0)$
Putting $x - x_0=s$ we get
$\text{v}^2+-\text{u}^2=2\text{as}\ \dots(\text{iii})$
The above three laws are valid under the conditions, only when the acceleration is uniform.
- Velocity$-$time relation : Let $dv$ be the change in velocity in time interval. dt. Then acceleration
$\text{dv}=\text{a dt}$
Integrating from $0 \rightarrow t$ when velocity changes from $u \rightarrow v$
$\int\limits^\text{v}_\text{u}\text{dv}=\text{a}\int\limits^{\text{t}}_\text{0}\text{dt}$
$\text{v}-\text{u}=\text{at}$
$\text{v}=\text{u}+\text{at}\ \dots(\text{i})$
- Distance$-$time relation: Consider an object moving in a straight line with uniform acceleration $'a\ '$. Let at any instant $t, dx$ be the displacement of the object in time interval at., Then instantaneous velocity $v$ is given by:
$\text{dx}=\text{vdt}$
$\text{dx}=(\text{u}+\text{at})\text{dt} [$from $(i) \text{v}=\text{u}+\text{at}]$
Let $x_0 =$ displacement at $t = 0$
$x =$ displacement at $t = t$
Integrating within limits
$\int\limits^\text{x}_{\text{x}_0}\text{dx}=\int\limits^{\text{t}}_0(\text{u}+\text{at})\text{dt}$
$=\text{u}\int\limits^{\text{t}}_0\text{dt}+\text{a}\int\limits^{\text{t}}\text{tdt}$
$\text{x}-\text{x}_0=\text{ut}+\frac{1}{2}\text{at}^2$
$\text{x}=\text{x}_0+\text{ut}+\frac{1}{2}\text{at}^2\ \dots(\text{ii})$
If $x - x_0 = s$ = distance covered by an object in time $t$ then
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
- Velocity-displacement relation.
then $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{dv}}{\text{dx}}\times\frac{\text{dx}}{\text{dt}}=\text{v}\frac{\text{dv}}{\text{dx}}$
$\text{adx}=\text{vdv}$
Let $u$ be the velocity of object at position $x_0\ v$ be the velocity of object at position $x$
Integrating above within limits
$\int\limits^{\text{x}}_{\text{x}_0}\text{a dx} = \int\limits^{\text{c}}_{\text{u}}\text{v dv}$
$\text{a}(\text{x}-\text{x}_0)=\frac{\text{v}^2}{2}-\frac{\text{u}^2}{2}$
$\text{v}^2-\text{u}^2=2\text{a}(\text{x}-\text{x}_0)$
Putting $x - x_0=s$ we get
$\text{v}^2+-\text{u}^2=2\text{as}\ \dots(\text{iii})$
The above three laws are valid under the conditions, only when the acceleration is uniform.
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