# Question #1d51d

##### 1 Answer

#### Answer:

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

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

BD = BC – CD, is the change in velocity during time

Draw AD parallel to OC.

In the figure

BC = BD + DC = BD + OA

Now BC =

Therefore we get

#v# = BD +#u#

#=># BD =#v - u# ........(1)

We know that rate of change of velocity is acceleration

#:.a = ("change in velocity")/("time taken")#

#=>a= "BD"/"AD" = "BD"/"OC"#

Since OC =

#a = "BD"/ t#

#=> "BD" = at# .............(2)

From (1) and (2) we get

#v = u + at# .......(3)

From the Fig. the distance traveled

Now

Area of OABC

Substituting OA =

#=>s = u t + 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

Area of the trapezium OABC

#= ("OA" + "BC")/2 xx# OC

Substituting OA =

#s = (u + v)/2 xxt # .......(5)

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

#t = (v - u)/a# ...........(6)

Substituting value of

#s = (v + u)/2 xx (v - u) /a#

#=>2 a s = v^2 – u^2#