Kinematics>>Equations of Motion

[SunnyH]

Homework Statement

A motorcyclist is traveling at 15.0m/s [forward] and applies brakes. The motorcycle slows down at 5.0m/s [backward].
a) Determine the motorcycle's breaking distance [ans:23 m [forward]]
Given:
Vinitial: 15.0m/s [forward]
a: 5.0m/s^2 [backward]
Vfinal: 0 m/s [forward]

Homework Equations

d= vi(T) + 1/2a(T^2) >>> vector equation
vf = vi +aT >>>> vector equation

The Attempt at a Solution

1) Find time for breaking distance
vf = vi + aT
0 = vi - aT (making acceleration negative so that it's vector is forward)
0 = 15 - 5T
-15/-5 =T
3 = T
2) Finding breaking distance
d= vi(T) + 1/2a(T^2)
d= vi(T) - 1/2a(T^2) (acceleration made forward by making it negative)
d = 15(3) - 2.5(9)
d = 75 - 22.5
d = 52.5 m
Therefore the braking distance is 52.5 m but the answer is 23m which I think came by ignoring initial velocity but it doesn't make any sense to me why they ignore initial velocity so kindly explain that to me. The picture i uploaded explains why i think initial velocity must be considered.

 

[kuruman]

d = 15(3) - 2.5(9)
d = 75 - 22.5

15(3) = 45 not 75.