V= at or velocity = acceleration x time square

[sweet_gen03]

V= at or velocity = acceleration x time square

so how did they arrived like this formula?




thank you for your reply

need help ,

gen

 

[arunbg]

The correct relation is v = u + at or a = (v-u)/t
where u stands for initial velocity, v final velocity, and a the [B}constant linear[/B] acceleration.
The result follows since acceleration is defined to be the time rate of change of velocity, and since this rate is a constant here, you can simply take the ratio of the difference between any two velocities at different times, to the time taken . Hence the second equation.
Or simply,
$$a=\frac{dv}{dt}$$
$$\int adt=\int dv$$
$$v=at+C$$
where C is the constant of integration.
Putting t=0 in the expression, we get v=c=u.
The result follows.

 

[roundedge]

hooray for calculus

 

[HallsofIvy]

For the simple "constant acceleration" case, acceleration is defined as "change in speed"/"change in time". From that you immediately get "change in speed"= acceleration*"change in time" by multiplying both sides of the equation by "change in time".

Notice that the equation you give: v= at is incorrect in general. Writing $\Delta v$ for "change in speed", if t represents the length of time accelerating, then $\Delta v$= at. $\Delta v= v$ only if the initial velocity is 0. If the initial speed is v₀ we can get the final speed by adding "change in speed" to "initial speed":
v= at+ v₀.

If the acceleration is not constant, then "change in speed" divided by "change in time" only gives average acceleration. The acceleration at each time is "derivative of speed function with respect to time":
[tex]a(t)= \frac{dv(t)}{dt}[/itex]
and we go from acceleration to speed by integrating:
[tex]v(t)= \int_{t_0}^t a(t)dt[/itex]