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Ray Of Light [21]
3 years ago
12

Which describes the average velocity of an ant traveling at a constant speed

Physics
1 answer:
yanalaym [24]3 years ago
3 0

Answer: A. The total displacement divided by the time and  C. The slope of the ant's displacement vs. time graph.

Explanation:

Hi! The question seems incomplete, but I found the options on the internt:

A. The total displacement divided by the time.

B. The slope of the ant's acceleration vs. time graph.

C. The slope of the ant's displacement vs. time graph.

D. The average acceleration divided by the time.

Now, since we know the ant is travelling at a constant speed, its average velocity V will be expressed by the following equation:

V=\frac{d}{t}

Where:

d is the ant's total displacement

t is the time it took to the ant to travel to the kitchen

Hence one of the correct options is: A. The total displacement divided by the time

On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is: C. The slope of the ant's displacement vs. time graph.

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While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at
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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

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