A child slides down a hill on a toboggan with an acceleration of 1.5 m/s squared. If she starts at rest, how far has she traveled in 1.0 s, 2.0 s, and 3.0 s?

d = (1/2) a t^2

(1/2) a = .75
so
d = .75 t^2
for example if t = 2
d = .75 * 4 = 3m

To determine the distance traveled by the child on the toboggan, you can use the equations of motion. In this case, the child starts at rest, so her initial velocity (u) is 0 m/s. The acceleration (a) is given as 1.5 m/s².

The equation to compute distance (s) as a function of time (t), initial velocity (u), and acceleration (a) is:

s = ut + (1/2)at²

Let's calculate the distances for the given times:

For t = 1.0 s:
s₁ = (0)(1.0) + (1/2)(1.5)(1.0)²
s₁ = 0 + (1/2)(1.5)(1.0)
s₁ = 0 + 0.75
s₁ = 0.75 m

For t = 2.0 s:
s₂ = (0)(2.0) + (1/2)(1.5)(2.0)²
s₂ = 0 + (1/2)(1.5)(4.0)
s₂ = 0 + 3.0
s₂ = 3.0 m

For t = 3.0 s:
s₃ = (0)(3.0) + (1/2)(1.5)(3.0)²
s₃ = 0 + (1/2)(1.5)(9.0)
s₃ = 0 + 6.75
s₃ = 6.75 m

Therefore, the child has traveled 0.75 m in 1.0 second, 3.0 m in 2.0 seconds, and 6.75 m in 3.0 seconds while sliding down the hill on the toboggan.