Question

Question #68199

Answer 1

Well where i come from this equation is written as ;

`s = ut - 1/2 at^2`[ Deceleration]

Okay for this we use the 1st kinematics equation;

`v = u +at`

`s = ½ (u+v) t`

`s = ½ (u + v) × t`

`s = ½ (u + u + at) × t`

`s = ut + ½ at²` {here `+ at` as the acceleration is positive}

Using the 2 equations we already know we can derive the 3rd equation

Answer 2

Have a look:

I used `t_1` to represent reaction time instead of `t` in my deduction.