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

gravity

explanation:

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