Question

Question #1d51d

Answer 1

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`.

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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`