A physics question please help

Determine the stopping distances for a car with an initial speed of 89 km/h and human reaction time of 3.0 s for the following accelerations.

a) a = -4.0 m/s2

b) a = -8.0 m/s2

please help thanks!!

89 kph = 24.7 m/s

a) v^2 = u^2 – 2 .a . d

d = u^2 / 2 . a = 24.7 ^2 / 2 . 4 = 76.4 m

Total stopping distance = 78.4 + ( 3 . 24.7 ) = 150.5 m

b) d = 38.2

Total distance = 38.2 + 74.1 = 112.3 m

That 3 seconds is a very slow reaction time.

change the initial speed to m/s:

89km/h = 89000m/h = 24.72 m/s

the distance traveled before the brakes are applied is 24.72 m/s * 3 sec = 74.17 m

distance traveled during negative acceleration is computed by manipulating the formula:

vf^2 = vi^2 + 2*a*d and since vf (final velocity) is 0 then 0 = vi^2 + 2*a*d -vi^2 = 2*a*d so then

d = -vi^2/2a

for example a) d = -24.72^2/2*-4 d = - (-76.38) or 76.38m

total distance traveled = 76.38 + 74.17 = 150.55m

for example b) d = - 24.72^2/2*-8 d = -(-38.19) or 38.19m

total distance traveled is 38.19 + 74.17 = 112.36m

hello Amit Patel,
first it must to convert from Km/s to m/s
89 (km/h ) * 1000(m/1km) * 1(h/3600sec)
the result is (89 * 1000) /(3600) = 24.72 m/s
a=(vf - vi) / (tf - ti) " vf is the final speed while vi is the initial speed " "tf is the final time while fi is the initial"
(0 - 24.72) / ( 3 -0 ) = ( -24.72) / ( 3 ) = - 8 m/s^2
so the answer is b) a = - 8.0 m/s2