Question
State the kinematic equations for uniformly accelerated motion.
✓
Answer
For uniformly accelerated motion, we can derive some simple equations that relate displacement $(x)$, time taken $(t)$, Initial velocity $(u)$, final velocity $(v)$ and acceleration $(a)$.
Let $u$ be the initial velocity of the particle at $t = 0$ and $v$ is the final velocity of the particle after time $t$. Consider two points $A$ and $B$ on the curve corresponding to $t = 0$ and $t = t$ respectively.
Draw $BD$ perpendicular to time axis. Also draw $AC$ perpendicular to $BD$.
$\therefore OA = CD = u;$
$BC = (v - u)$ and $OD = t$
Now slope of $v-t$ graph $=$ acceleration $(a)$
$\therefore a =$ slope of $v-t$ graph $=\tan\theta=\frac{\text{BC}}{\text{AC}}=\frac{\text{BC}}{\text{OD}} [\because AC= OD]$
$\therefore \text{a}=\frac{\text{v}-\text{u}}{\text{t}}$
$\text{v}-\text{u}=\text{at}$
or $\text{v}=\text{u}+\text{at}$
$x =$ position of the particle at $t = t$ from the origin.
$\therefore (x - x_0) = S =$ distance travelled by the particle in the time interval $(t - 0) = t$
We know, distance travelled by a particle in the given time
Interval $=$ area under velocity$-$time graph
$\therefore (x – x_0) =$ Area $\text{OABD} ($see fig, above$) =$ Area of trapezium $\text{OABD}$
$=\frac{1}{2}[$ Sum of parallel sides$ \times$ perpendicular distance between parallel sides$]$
$=\frac{1}{2}(\text{OA}+\text{BD})\times\text{AC}=\frac{1}{2}(\text{u}+\text{v})\times\text{t}$
Since $\text{v}=\text{u}+\text{at}$
$\therefore(\text{x}-\text{x}_0)=\frac{1}{2}(\text{u}+\text{u}+\text{at})\times\text{t}$
$=\frac{1}{2}(2\text{u}+\text{at})\times\text{t}=\text{ut}+\frac{1}{2}\text{at}^2$
Since $\text{x}-\text{x}_0=\text{S}$
$\therefore \text{S}=\text{ut}+\frac{1}{2}\text{at}^2$
$S =$ area of trapenium $\text{OABD}$
$=\frac{1}{2} ($sum of parallel sides$) \times$ perpendicular distance between these para
$\text{S}=\frac{1}{2}(\text{OA}+\text{BD})\times\text{ACs}\ \dots(\text{i})$
Now, acceleratiory $a =$ slope of $v- t$ graph
$\text{a}=\frac{\text{BC}}{\text{AC}}=\frac{\text{BD}-\text{CD}}{\text{AC}}=\frac{\text{v}-\text{u}}{\text{AC}}$
$\text{AC}=\Big(\frac{\text{v}-\text{u}}{\text{a}}\Big)\text{s}$
Also, $OA = u$ and $BD = v$
Using equations $(ii)$ and $(iii)$ in equation $(i)$, we get
$\text{S}=\frac{1}{2}(\text{v}+\text{u})\frac{(\text{v}-\text{u})}{\text{a}}=\frac{\text{v}^2-\text{u}^2}{2\text{a}}$
$\text{v}^2-\text{u}^2=2\text{as}$
- Velocity attained after time $t$: The velocity$-$time graph for positive constant acceleration of a particle is shown in the figure.
Let $u$ be the initial velocity of the particle at $t = 0$ and $v$ is the final velocity of the particle after time $t$. Consider two points $A$ and $B$ on the curve corresponding to $t = 0$ and $t = t$ respectively.
Draw $BD$ perpendicular to time axis. Also draw $AC$ perpendicular to $BD$.
$\therefore OA = CD = u;$
$BC = (v - u)$ and $OD = t$
Now slope of $v-t$ graph $=$ acceleration $(a)$
$\therefore a =$ slope of $v-t$ graph $=\tan\theta=\frac{\text{BC}}{\text{AC}}=\frac{\text{BC}}{\text{OD}} [\because AC= OD]$
$\therefore \text{a}=\frac{\text{v}-\text{u}}{\text{t}}$
$\text{v}-\text{u}=\text{at}$
or $\text{v}=\text{u}+\text{at}$
- Distance travelled in time $t$:
$x =$ position of the particle at $t = t$ from the origin.
$\therefore (x - x_0) = S =$ distance travelled by the particle in the time interval $(t - 0) = t$
We know, distance travelled by a particle in the given time
Interval $=$ area under velocity$-$time graph
$\therefore (x – x_0) =$ Area $\text{OABD} ($see fig, above$) =$ Area of trapezium $\text{OABD}$
$=\frac{1}{2}[$ Sum of parallel sides$ \times$ perpendicular distance between parallel sides$]$
$=\frac{1}{2}(\text{OA}+\text{BD})\times\text{AC}=\frac{1}{2}(\text{u}+\text{v})\times\text{t}$
Since $\text{v}=\text{u}+\text{at}$
$\therefore(\text{x}-\text{x}_0)=\frac{1}{2}(\text{u}+\text{u}+\text{at})\times\text{t}$
$=\frac{1}{2}(2\text{u}+\text{at})\times\text{t}=\text{ut}+\frac{1}{2}\text{at}^2$
Since $\text{x}-\text{x}_0=\text{S}$
$\therefore \text{S}=\text{ut}+\frac{1}{2}\text{at}^2$
- Velocity attained after travelling a distance $S$: We know, distance travelled by a particle in time $t$ is equal to the area under velocity$-$time graph. Therefore, the distance $(s)$ travelled by a particle during time interval $t$ is given by
$S =$ area of trapenium $\text{OABD}$
$=\frac{1}{2} ($sum of parallel sides$) \times$ perpendicular distance between these para
$\text{S}=\frac{1}{2}(\text{OA}+\text{BD})\times\text{ACs}\ \dots(\text{i})$
Now, acceleratiory $a =$ slope of $v- t$ graph
$\text{a}=\frac{\text{BC}}{\text{AC}}=\frac{\text{BD}-\text{CD}}{\text{AC}}=\frac{\text{v}-\text{u}}{\text{AC}}$
$\text{AC}=\Big(\frac{\text{v}-\text{u}}{\text{a}}\Big)\text{s}$
Also, $OA = u$ and $BD = v$
Using equations $(ii)$ and $(iii)$ in equation $(i)$, we get
$\text{S}=\frac{1}{2}(\text{v}+\text{u})\frac{(\text{v}-\text{u})}{\text{a}}=\frac{\text{v}^2-\text{u}^2}{2\text{a}}$
$\text{v}^2-\text{u}^2=2\text{as}$
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