For uniformly accelerated motion, we can derive some simple equations that relate displacement (`x`), time taken (`t`), initial velocity (`v_0`), final velocity (`v`) and acceleration (`a`).
`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "First Equation")} `
If the velocity of an object is `v_o` at `t = 0` and `v` at time `t`, we have
`color(blue)(bara=(v-v_o)/(t-0))`
`color(blue)(v=v_o +at)`...................................`(i)`
`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Second Equation")} `
The area under this curve is :
Area between instants 0 and t = Area of `"triangle" ABC +` Area of `"rectangle" OACD`
`=1/2(v-v_o)t + v_o t`
`color(green)("We know, the area under v-t curve represents the displacement.")` Therefore,
`color(green)(x=1/2(v-v_o)t + v_o t)`
But `v-v_0 = a t` Therefore, `x=1/2 at^2 +v_ot`
`color(blue)(x=v_ot+1/2at^2)`...........................................`(ii)`
`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Third Equation")} `
As we discussed earlier, `x=1/2(v-v_o)t + v_o t`
This equation can also be written as
`x=(v+v_0)/2 t=barv t`.......(a)
where, `barv=(v+v_o)/2`
From first kinematic equation `t=(v-v_o)/a`
Substituting this in Eq. (a), we get
`x=barv t=((v+v_o)/2)((v-v_o)/a)=(v^2-v_o^2)/2`
`color(blue)(v^2=v_o^2 +2ax)`..........................................`(iii)`
Thus, `ul"we have obtained three important equations :"`
`color(blue)(v=v_o +at)`
`color(blue)(x=v_ot+1/2at^2)`
`color(blue)(v^2=v_o^2 +2ax)`
For uniformly accelerated motion, we can derive some simple equations that relate displacement (`x`), time taken (`t`), initial velocity (`v_0`), final velocity (`v`) and acceleration (`a`).
`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "First Equation")} `
If the velocity of an object is `v_o` at `t = 0` and `v` at time `t`, we have
`color(blue)(bara=(v-v_o)/(t-0))`
`color(blue)(v=v_o +at)`...................................`(i)`
`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Second Equation")} `
The area under this curve is :
Area between instants 0 and t = Area of `"triangle" ABC +` Area of `"rectangle" OACD`
`=1/2(v-v_o)t + v_o t`
`color(green)("We know, the area under v-t curve represents the displacement.")` Therefore,
`color(green)(x=1/2(v-v_o)t + v_o t)`
But `v-v_0 = a t` Therefore, `x=1/2 at^2 +v_ot`
`color(blue)(x=v_ot+1/2at^2)`...........................................`(ii)`
`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Third Equation")} `
As we discussed earlier, `x=1/2(v-v_o)t + v_o t`
This equation can also be written as
`x=(v+v_0)/2 t=barv t`.......(a)
where, `barv=(v+v_o)/2`
From first kinematic equation `t=(v-v_o)/a`
Substituting this in Eq. (a), we get
`x=barv t=((v+v_o)/2)((v-v_o)/a)=(v^2-v_o^2)/2`
`color(blue)(v^2=v_o^2 +2ax)`..........................................`(iii)`
Thus, `ul"we have obtained three important equations :"`
`color(blue)(v=v_o +at)`
`color(blue)(x=v_ot+1/2at^2)`
`color(blue)(v^2=v_o^2 +2ax)`