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!!

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

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

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

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

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

thanks a million for ur time and effort!!

To determine the stopping distances for an automobile, we will use one of Newton's equations of motion, specifically the one that relates distance, initial velocity, acceleration, and time. The equation we will use is:

distance = (initial velocity * time) + (0.5 * acceleration * time^2)

Let's plug in the values given in the question:

(a) For an acceleration (a) of -40m/s^2 and a human reaction time (t) of 1.0s, the initial velocity (u) is 90km/h. We need to convert the initial velocity to meters per second before solving the equation.

1 km = 1000 m and 1 hour = 3600 s
So, 90 km/h = (90 * 1000) / 3600 = 25 m/s

Using the equation, we can now calculate the stopping distance:

distance = (25 * 1) + (0.5 * -40 * 1^2)
distance = 25 - 20
distance = 5 meters

Therefore, the stopping distance for an acceleration of -40m/s^2 is 5 meters.

(b) For an acceleration (a) of -8.0m/s^2 and a human reaction time (t) of 1.0s, the initial velocity (u) is still 25 m/s, as it was not mentioned otherwise.

Using the equation again, we can calculate the stopping distance:

distance = (25 * 1) + (0.5 * -8.0 * 1^2)
distance = 25 - 4
distance = 21 meters

Therefore, the stopping distance for an acceleration of -8.0m/s^2 is 21 meters.

To sum up, we used Newton's equation of motion, specifically the one that relates distance, initial velocity, acceleration, and time, to calculate the stopping distances for different accelerations given an initial velocity and human reaction time.