This question involves the concepts of the equations of motion, frictional force, Newton's Second Law and uniform speed.
(1) The acceleration of the car "- 9.62 m/s²".
(2) The braking distance of the car in front is "34.06 m".
(3) The minimum safe distance between the cars is "18.71 m".
(1)
The acceleration (deceleration) of the car can be calculated using the formula for frictional force:
[tex]F = -\mu mg[/tex] (negative sign due to opposite direction to motion)
using Newton's Second Law:
[tex]ma = -\mu mg\\a = -\mu g[/tex]
where,
a = acceleration = ?
μ = coefficcient of friction = 0.981
g = acceleation due to gravity = 9.81 m/s²
Therefore,
[tex]a = -(0.981)(9.81\ m/s^2)\\[/tex]
a = -9.62 m/s²
(2)
The braking distance can be found using the third equation of motion:
[tex]2as = v_f^2-v_i^2[/tex]
where,
s = braking distance = ?
vf = final speed = 0 m/s
vi = initial speed = 25.6 m/s
Therefore,
[tex]2(-9.62\ m/s^2)s=(0\ m/s)^2-(25.6\ m/s)^2\\\\s = \frac{-655.36\ m^2/s^2}{-19.24\ m/s^2}\\\\[/tex]
s = 34.06 m
(3)
The safe distance will be equal to the extra distance travelled by car before applying the brake, during the reaction time. Since it is the uniform speed motion, hence using the equation:
[tex]d = vt\\d = (25.6\ m/s)(0.731\ s)[/tex]
d = 18.71 m
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
