Question

x-xnaught=vnaught + (1/2)at^2
Naught is like x0 (starting position)
how was the above equation derived? Please explain in simplest terms. math proof not really necessary. but logic will suffice.

Answer 1

a is acceleration
v is velocity

Answer 2

actually vnaught (t) srry!

Answer 3

\(\large x - x_0 = v_0t + \frac{1}{2}at^2\)

Answer 4

right, ok

Answer 5

u hav some familiarity wid calculus ?

dx/dt, d^x/dt^2 etc...

Answer 6

a bit like derivatives, chain rules, product rule,

Answer 7

that will do :) lets analyze the given equation

Answer 8

it looks like a quadratic in t ?

Answer 9

\(\large x - x_0 = v_0t + \frac{1}{2}at^2\)

\(\large x = x_0 + v_0t + \frac{1}{2}at^2\)

Answer 10

notice that \(x\) is a quadratic function in \(t\),

Answer 11

u must be knowing that \(x\) is position of object, and \(t\) is time

Answer 12

lets take a simple quadratic equation and try to derive above

Answer 13

sorry my internet disconnected for a while!

Answer 14

its okay :)

Answer 15

we want formula for position of one-dimensional motion wid constant acceleration.

so we better take a quadratc :-

\(\large x = at^2 + bt + c\)

Answer 16

thats the position function.
how do we get VELOCITY function from above ?

Answer 17

\(\large v = \frac{dx}{dt}\)

right ?

so simply take derivative of position function , \(x\)

Answer 18

\(\large v = \frac{dx}{dt} = \frac{d}{dt} (at^2+ bt + c)\)

Answer 19

wat do u get ?

Answer 20

2at+b right?

Answer 21

exactly, thats the velocity of object

Answer 22

y-yo=1/2(a)(t-to)^2+vo(t-to)

Answer 23

\(\large x = at^2+ bt + c\) --------(1)


\(\large v = \frac{dx}{dt} = \frac{d}{dt} (at^2+ bt + c) = 2at + b\) ----(2)

Answer 24

find acceleration now

Answer 25

\(\large a = \frac{dv}{dt} = \frac{d}{dt} (2at+b) = ?\)

Answer 26

that would be the derivative of 2at+b which is 2a??

Answer 27

yup !

Answer 28

\(\large x = at^2+ bt + c\) --------(1)

\(\large v = \frac{dx}{dt} = \frac{d}{dt} (at^2+ bt + c) = 2at + b\) ----(2)

\(\large acc = \frac{dv}{dt} = \frac{d}{dt} (2at+b) = 2a\) ----(3)

Answer 29

from equation (3), can u solve the coefficient a ?

Answer 30

before u use the equation u need to think about position and time that it start

Answer 31

from (3)

\(\large acc = 2a\)


=>?

\(\large a = \frac{acc}{2}\) ----------------(4)

Answer 32

can u find the remaining coefficients, b and c

Answer 33

once we have values for , a, b and c,
we can simplyy substitute this in (1) and get the equation for one-dimensional motion wid constant acc

Answer 34

thats a good point :) lets get to that after finishing playing wid quadratic @sovannayen

Answer 35

b=v-At
and c .. this is sort of complicated..

Answer 36

yes, put \(t=0\), then \(b\) becomes velocity at t=0 eh ?

Answer 37

why is t 0?

Answer 38

put t= 0 in equation (2),

\(\large v_0 = b\)

Answer 39

because, we want to derive final equaiton in terms of \(v_0\), which is the velocity at t=0

Answer 40

oh I see..

Answer 41

im sure that doesnt convince u much, let me think of explaining u better

Answer 42

or, u can help me lol why t=0 above ?

Answer 43

no I think I actually do understand. v naught equals the velocity at starting time. except, there should be a velocity at starting time shouldn't there?

Answer 44

*shouldn't

Answer 45

there will be starting velocity for any object,

Answer 46

starting velocity can be 0, when its freeling falling from a height,

Answer 47

but I don't think there is a velocity, in other words, there is a 0 velocity at starting time right?

Answer 48

but when u throw a stone, it surely will have a starting velocity other than 0