a rollercoaster car is moving 19.8 m/s on flat ground when it's hits the brakes. it decelerates at - 3.77 m/s^2 over the next 45.8 m. how much time does it take?

3.44 s

Explanation:

Given:

Δx = 45.8 m

v₀ = 19.8 m/s

a = - 3.77 m/s²

Find: t

Δx = v₀ t + ½ at²

45.8 = 19.8 t + ½ (-3.77) t²

45.8 = 19.8 t - 1.885 t²

1.885 t² - 19.8 t + 45.8 = 0

t = [ - b ± √ (b² - 4ac) ] / 2a

t = [ 19.8 ± √ ((-19.8) ² - 4 (1.885) (45.8)) ] / 2 (1.885)

t = [ 19.8 ± √ (392.04 - 345.332) ] / 3.77

t = 3.44 or 7.06

The first time, 3.44 seconds, is correct.

The second time is the time it would take the car to return to 45.8 m if it continued decelerating after stopping.