Acar traveling 77 km/h slows down at a constant 0.45 m/s^2 just by "letting up on the gas." --part a : calculate the distance the car coasts before it stops.
--part b: calculate the time it takes to stop.
--part c: calculate the distance it travels during the first second.
--part d: calculate the distance it travels during the fifth second. need with all parts a-d, show all work and formulas used.

(a) 508.37 m

(b) 47.53 s

(c) 21.165 m

(d) 19.365 m

Explanation:

initial velocity, u = 77 km/h = 21.39 m/s

acceleration, a = - 0.45 m/s^2

(a) final velocity, v = 0

Let the distance traveled is s.

Use third equation of motion

s = 508.37 m

(b) Let t be the time taken to stop.

Use first equation of motion

v = u + at

0 = 21.39 - 0.45 t

t = 47.53 s

(c) Use the formula for the distance traveled in nth second

where n be the number of second, a be the acceleration, u be the initial velocity.

put n = 1, u = 21.39 m/s , a = - 0.45m/s^2

(d)  Use the formula for the distance traveled in nth second

where n be the number of second, a be the acceleration, u be the initial velocity.

put n = 5, u = 21.39 m/s , a = - 0.45m/s^2

d eras and periods

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!