Question

A toy 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?

Answer 1

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.