Atoy car goes over a small ramp at a horizontal velocity of 1.21 m/s and decelerates at 0.131 m/s2 while in the air. the total time elapsed was 0.342 seconds. what was the horizontal distance traveled?

We need to considerate only the horizontal component of the motion of the toy car.

The formula for the distance in a decelerated motion is:
s = s₀ + v₀·t - 1/2·a·t²

where:
s₀ = initial position = 0
v₀ = initial velocity = 1.21 m/s
t = time elapsed = 0.342 s
a = deceleration = 0.131 m/s²

Plugging in numbers:
s = 0 + 1.21×0.342 - 0.5×0.141×(0.342)²
  = 0.406 m

Hence, the toy car traveled a distance of about 41 cm.

Well the magnitude of its acceleration related to that car starts with the oil change like when you have to change the oil in your car you have to make sure that your precious car is starting really propley and very well

Ithink it's  c. i'm sure