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?

Question

Physics

Chuckles

10 January, 05:30

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?

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Answers (1)

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Konner Salinas · 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.