Since the motor cycle was moving at 36 m/s before the brakes were applied, the initial velocity
[tex]u = 36 {ms}^{ - 1} [/tex]
The motorcycle traveled 23m before stopping, that means
[tex]s = 23m[/tex]
Since the motorcycle stopped moving the final velocity is zero.
[tex]v = 0{ms}^{ - 1} [/tex]
We can use this 'suvat' equation of linear motion.
[tex] {v}^{2} = {u}^{2} + 2as[/tex]
Plugging in the above values gives,
[tex] {0}^{2} = {36}^{2} + 2a(23)[/tex]
[tex] 0 = 1296+ 46a[/tex]
[tex]46a = - 1296[/tex]
[tex]a = - 28.2 {ms}^{ - 2} [/tex]
We can use the equation
[tex]v = u + at[/tex]
to find the time the motorcycle took to stop.
[tex]0 = 36 + - 28.2t[/tex]
[tex] - 36 = - 28.2t[/tex]
[tex]t = 1.3s[/tex]
