Question

determine the stopping distances for an auto0mobile with the initial speed of 90km/hand the human reaction time of 1.0s: (a) for an acceleration a= -40m/s^2; (b) for a = -8.0m/s^2

use one of newtons four equations to solve....

i don't get this at all...PLEASE HELP!!

Answer 1

how far does the car go in that first second before braking?
90 *10^3 m/3600 s * 1 s = 25 meters bfore the brake is pushed.

then how far before stopping with initial speed of 25 m/s and acceleration of -40 m/s^2? (that is 4 times the acceleration of gravity - impossible brakes, probably a typo and they mean -4 m/s^2 but doing it anyway)

v = Vo + a t
0 = 25 -40 t
t = .625 seconds

d = Vo t + (1/2) a t^2
d = 25 (.625) - 20 (.625)^2
= 7.81 m
25 + .781 = 32.8 m

do the -8m/s^2 the same way

Answer 2

actually it is -4. it was actually meant to say -4.0, but i just forgot to put the decimal

Answer 3

and how did u get t=.625 seconds???

Answer 4

ok, just use -4 instead of -40'

v = Vo + a t
Vo = 25 m/s
when the thing stops, v = 0
so time to stop
0 = 25 + a t
NOW we have a = -4
0 = 25 -4 t
4 = 25/4 = 6.25 seconds to stop after the brakes are applied

Answer 5

so
d = 25 (6.25) - 2 (6.25)^2
= 156.25 - 78.125
= 78.125 meters after the brakes are applied so
25 + 78.125 = 103 meters total

Now repeat with -8 m/s^2 instead of -4 m/s^2

Answer 6

thanks a million for ur time and effort!!