Question #1d51d

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#.

Draw AD parallel to OC.

In the figure
BC = BD + DC = BD + OA
Now BC = #v# and OA = #u#

Therefore we get

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

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

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

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

Since OC = #t# we get

#a = "BD"/ t#
#=> "BD" = at# .............(2)

From (1) and (2) we get

#v = u + at# .......(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
#:.s=# OA #xx# OC + #1/2# (AD #xx# BD) .......(4)

Substituting OA = #u#, OC = AD = #t and# BD = #at#, in (4) we get
#s = u xx t + 1/2 (txxat )#

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

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

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

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

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

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

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