Question

A cyclist is travelling at 15 m|s-1. She applies brakes so that she could not collide with 18 m away what deceleration must she have?

Answer 1

S-U-V-A-T-- S-displacement, u-initial
velocity, v-final velocity,acceleration,
t-time
v2 = u2 +2as
s=1/2(utv)ts=ut + 1/2at2

Her initial velocity, u is 15m/s
Her final velocity is, v Om/s (since she
stops)
Displacement is, s 18m
Hence the acceleration, a:

v2=u2 + 2as

0=15^2 + 2 xa x 18
0=225 + 36a
36a= -225

a= -225/36
a= -6.25 m/s/s

Therefore she has to decelerate at the
rate of 6.25m/s2

Answer 2

u=15m/s
v=0m/s
$2as= v^{2} - u^{2}$
$2a*18= 0^{2}- 15^{2}$[tex]36a=-225 \\ a= \frac{-225}{36} \\ a=-6.25 [/tex]