Answer:
The speed of the moving walkway is 1.50 m/s
Explanation:
The position of the child can be calculated using the following equation:
x = x0 + v · t
Where :
x = position of the child at time t.
v = velocity of the child.
t = time.
When the child runs in the same direction as the walkway, the velocity of the child will be its velocity relative to the walkway plus the velocity of the walkway. Then, if we place the origin of the frame of reference at the start of the walkway:
x = x0 + v · t
25 m = 0 m + (2.8 m/s + v) · t₁
Where v is the velocity of the walkway
On its way back, the velocity of the child relative to the walkway is in the opposite direction to the velocity of the walkway. Then:
x = x0 + v · t
0 m = 25 m + (-2.8 m/s + v) · t₂
We also know that t₁ + t₂ = 25 s
Then: t₁ = 25 - t₂
So, we can write the following system of equations:
25 m = (2.8 m/s + v) · (25 s - t₂)
-25 m = (-2.8 m/s + v) · t₂
Let´s take the second equation and solve it for t₂
-25 m / (-2.8 m/s + v) = t₂
Now, let´s replace t₂ in the first equation:
25 m = (2.8 m/s + v) · (25 s + 25 m / (-2.8 m/s + v))
Let´s sum the fraction: 25 s + 25 m / (-2.8 m/s + v)
25 m = (2.8 m/s + v) · (25 s ·(-2.8 m/s + v) + 25 m) / (-2.8 m/s + v)
multiply by (-2.8 m/s + v) both sides of the equation:
25 m(-2.8 m/s + v) = (2.8 m/s + v) · (-70 m + 25 s · v + 25 m)
Apply distributive property:
-70 m²/s +25 m·v = -196 m²/s +70 m·v +70 m²/s -70 m·v +25 s ·v² + 25 m v
56 m²/s = 25 s · v²
56 m²/s / 25 s = v²
v = 1.50 m/s
The speed of the moving walkway is 1.50 m/s
