Question

Derive the equation velocity time relation from velocity time graph

Answer 1

Explanation:

Let an object is moving with uniform acceleration.

Velocity Time Graph4

Let an object is moving with uniform acceleration.

Let the initial velocity of the object = u Let the object is moving with uniform acceleration, a. Let object reaches at point B after time, t and its final velocity becomes, v Draw a line parallel to x-axis DA from point, D from where object starts moving. Draw another line BA from point B parallel to y-axis which meets at E at y-axis.

Let OE = time,

t Now,

from the graph,

BE = AB + AE

⇒ v = DC + OD

(Since, AB = DC and AE = OD)

⇒ v = DC + u (Since, OD = u)

⇒ v = DC + u ------------------- (i)

Now,

Acceleration (a) =(Change in velocity)/(Time taken)

a= v – ut

⇒ a = v - ut

⇒ a= OC – OD t = DC t

⇒a = OC-OD t =DC t

⇒at=DC

⇒at=DC -----(ii)

By substituting the value of DC from (ii) in (i),

we get v= at + u v = at + u

⇒ v = u + at

⇒ v = u + at

Above equation is the relation among initial velocity (u), final velocity (v), acceleration (a) and time (t).

It is called first equation of motion.

Answer 2

ᐯᗴᒪOᑕITY-TIᗰᗴ ᘜᖇᗩᑭᕼ: ՏᕼOᗯՏ Tᕼᗴ ᑕᕼᗩᑎᘜᗴ Iᑎ ᐯᗴᒪOᑕITY ᗯITᕼ TIᗰᗴ Oᖴ ᗩ ᑌᑎIᖴOᖇᗰᒪY ᗩᑕᑕᗴᒪᗴᖇᗩTᗴᗪ OᗷᒍᗴᑕT. Tᕼᗴ OᗷᒍᗴᑕT ՏTᗩᖇTՏ ᖴᖇOᗰ Tᕼᗴ ᑭOIᑎT ᗪ Iᑎ Tᕼᗴ ᘜᖇᗩᑭᕼ ᗯITᕼ ᐯᗴᒪOᑕITY ᐯ. ITՏ ᐯᗴᒪOᑕITY KᗴᗴᑭՏ IᑎᑕᖇᗴᗩՏIᑎᘜ ᗩᑎᗪ ᗩᖴTᗴᖇ TIᗰᗴ T. IT ᖇᗴᗩᑕᕼᗴՏ Tᕼᗴ ᑭOIᑎT ᗷ Oᑎ Tᕼᗴ ᘜᖇᗩᑭᕼ.

Tᕼᗴ IᑎITIᗩᒪ ᐯᗴᒪOᑕITY Oᖴ Tᕼᗴ OᗷᒍᗴᑕT= ᑌ= Oᗪ

Tᕼᗴ ᖴIᑎᗩᒪ ᐯᗴᒪOᑕITY Oᖴ Tᕼᗴ OᗷᒍᗴᑕT= ᐯ= Oᑕ

TIᗰᗴ= T= Oᗴ

ᗩᑕᑕᗴᒪᗴᖇᗩTIOᑎ (ᗩ)= ᑕᕼᗩᑎᘜᗴ Iᑎ ᐯᗴᒪOᑕITY/ TIᗰᗴ

= (ᖴIᑎᗩᒪ ᐯᗴᒪOᑕITY- IᑎITIᗩᒪ ᐯᗴᒪOᑕITY)/ TIᗰᗴ

= (Oᑕ- Oᗪ)/ T

ᑕᗪ= ᗩT ....(1) (Oᑕ- Oᗪ= ᑕᗪ)

ᖴᖇOᗰ Tᕼᗴ ᘜᖇᗩᑭᕼ ....ᗷᗴ= ᗩᗷ+ ᗩᗴ

ᐯ= ᑕᗪ+ Oᗪ .....(ᗩᗷ= ᑕᗪ ᗩᑎᗪ ᗩᗴ= Oᗪ)

ᐯ= ᗩT+ ᑌ .....(ᖴᖇOᗰ 1)

ᐯ= ᑌ+ ᗩT

TᕼIՏ IՏ Tᕼᗴ ᖴIᖇՏT ᗴᑫᑌᗩTIOᑎ Oᖴ ᗰOTIOᑎ.

ᗴᑫᑌᗩTIOᑎ ᗪᗴՏᑕᖇIᗷIᑎᘜ ᖇᗴᒪᗩTIOᑎ ᗷᗴTᗯᗴᗴᑎ ᐯᗴᒪOᑕITY & TIᗰᗴ.