Acar is able to stop with an acceleration of − 3.00 m/s^2. justify the mathematical routine used to calculate the distance required to stop from a velocity of 100.0 km/h by choosing the correct answer below.
a.) 16.7m because the average velocity is 50 km/h, the change in velocity divided by the acceleration is the time, and the time multiplied by the average velocity is the distance.
b.) 64.4m because the average velocity is 13.9 m/s, the average velocity divided by the acceleration is the time, and the time multiplied by the average velocity is the distance.
c.) 129m because the average velocity is 13.9 m/s, the change in velocity divided by the acceleration is the time, and the time multiplied by the average velocity is the distance.
d.) 257m because the initial velocity is 27.8 m/s, the initial velocity divided by the acceleration is the time, and the time multiplied by the initial velocity is the distance.

129 m because the average velocity is 13.9 m/s, the change in velocity

divided by the acceleration is the time, and the time multiplied by the

average velocity is the distance. ⇒ answer C

Explanation:

Lets explain how to solve the problem

The given is:

The care is able to stop with an acceleration of -3 m/s²

→ The final velocity = 0 and acceleration = -3 m/s²

Calculate the distance required to stop from a velocity of 100 km/h

→ Initial velocity = 100 km/h

At first we must to change the unite of the initial velocity from km/h

to m/s because the units of the acceleration is m/s²

→ 1 km = 1000 meters and 1 hr = 3600 seconds

→ 100 km/h = (100 × 1000) ÷ 3600 = 27.78 m/s

The initial velocity is 27.78 m/s

Acceleration is the rate of change of velocity during the time,

then the time is the change of velocity divided by the acceleration

→

where v is the final velocity, u is the initial velocity, t is the time and

a is the acceleration

→ v = 0 , u = 27.78 m/s , a = -3 m/s²

Substitute these values in the rule

→ seconds

The time to required stop is 9.26 seconds

We can calculate the distance by using the rule:

→ s = ut + at²

→ u = 27.78 m/s , t = 9.26 s , a = -3 m/s²

Substitute these values in the rule

→ s = 27.78(9.26) + (-3)(9.26) = 128.6 ≅ 129 m

The distance required to stop is 129 m

Average velocity is total distance divided by total time

→ Total distance = 129 m and total time = 9.26 s

→ average velocity = 129 ÷ 9.26 = 13.9 m/s

The average velocity is 13.9 m/s

So the time multiplied by the average velocity is the distance

The answer is C

129 m because the average velocity is 13.9 m/s, the change in

velocity divided by the acceleration is the time, and the time

by the average velocity is the distance.

the answer is 3.1 his answer is wrong mines is right