# Question 1d51d

Nov 1, 2017

Equations (3) and (5) are the required equations.

#### Explanation:

There are two kinematic equations connecting velocity and time.

Consider a velocity-time graph of an object which moves with uniform acceleration as shown in Fig below.

Object has initial velocity $u$ at point A, which increases to value $v$, point B, in time $t$. The velocity changes at a uniform rate $= a$.

We draw perpendicular lines BC and BE from point B on time and velocity axes respectively.

Initial velocity is represented by OA, the final velocity is represented by BC and the time interval $t$ is represented by OC.
BD = BC – CD, is the change in velocity during time $t$. In the figure
BC = BD + DC = BD + OA
Now BC = $v$ and OA = $u$

Therefore we get

$v$ = BD + $u$
$\implies$ BD = $v - u$ ........(1)

We know that rate of change of velocity is acceleration $a$

$\therefore a = \left(\text{change in velocity")/("time taken}\right)$

$\implies a = \text{BD"/"AD" = "BD"/"OC}$

Since OC = $t$ we get

$a = \frac{\text{BD}}{t}$
$\implies \text{BD} = a t$ .............(2)

From (1) and (2) we get

$v = u + a t$ .......(3)

From the Fig. the distance traveled $s$ by the object is the area of the trapezium OABC.

Now
Area of OABC$=$ area of the rectangle OADC + area of the $\Delta$ ABD
$\therefore s =$ OA $\times$ OC + $\frac{1}{2}$ (AD $\times$ BD) .......(4)

Substituting OA = $u$, OC = AD = $t \mathmr{and}$ BD = $a t$, in (4) we get
$s = u \times t + \frac{1}{2} \left(t \times a t\right)$

$\implies s = u t + \frac{1}{2} a {t}^{2}$ .......(5)

=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=

For sake of completeness
Kinematic Equation for Position-Velocity relation is established as follows.

In the velocity-time graph shown in the Fig. above the distance $s$ travelled by the object in time $t$ while moving under uniform acceleration $a$ is the area of the trapezium OABC. From geometry we know that

Area of the trapezium OABC$= \frac{\text{OA" + "BC}}{2} \times$ OC

Substituting OA = $u$, BC = $v$ and OC = $t$ we get

$s = \frac{u + v}{2} \times t$ .......(5)

The velocity-time relation (3) can be rewritten as

$t = \frac{v - u}{a}$ ...........(6)

Substituting value of $t$ from (6) in (5) we get

$s = \frac{v + u}{2} \times \frac{v - u}{a}$
=>2 a s = v^2 – u^2#