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.

50 inches should be the resultant vector

d.7.5cm is your answer