Physics worksheet focusing on displacement, velocity, and acceleration with real-world scenarios and calculations.

It looks like you've uploaded a worksheet titled "Displacement, Velocity and Acceleration Worksheet" with several physics problems. I'll solve each problem step by step and explain the reasoning.

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Problem 1:

> Mike is traveling along a straight interstate highway. He notices that the mile marker reads 260. 30 minutes later, he reaches the 350-mile marker and then returns to the path to the 175-mile marker. What is Mike's displacement from the 260-mile marker?

#### Solution:

- Initial position: 260 miles
- Final position: 175 miles

Displacement = Final position – Initial position

$$
\text{Displacement} = 175 - 260 = -85 \text{ miles}
$$

The negative sign indicates direction (backwards from the initial point).

✔ Answer: $ \boxed{-85} $ miles

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Problem 2:

> A physics book is moved across the perimeter of a table of dimensions 1.0 m by 2.0 m.
>
> a. If the book ends up in its initial position, what is its displacement?

#### Solution:

- Displacement is the straight-line change in position.
- Since it ends at the same point where it started, the net change in position is zero.

✔ Answer: $ \boxed{0 \text{ m}} $

> b. What is the distance traveled?

- Distance is the total path length.
- The book travels around the perimeter of the rectangle:
Perimeter = $ 2 \times (1.0 + 2.0) = 2 \times 3.0 = 6.0 $ meters

✔ Answer: $ \boxed{6.0 \text{ m}} $

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Problem 3:

> Light from the sun reaches the Earth in 8.3 minutes. The velocity of light is $ 3.00 \times 10^8 $ m/s. How far from Earth is the sun?

#### Solution:

Use the formula:
$$
\text{Distance} = \text{Velocity} \times \text{Time}
$$

Convert time to seconds:
$$
8.3 \text{ min} = 8.3 \times 60 = 498 \text{ s}
$$

Now calculate:
$$
\text{Distance} = (3.00 \times 10^8 \, \text{m/s}) \times (498 \, \text{s}) = 1.494 \times 10^{11} \, \text{m}
$$

✔ Answer: $ \boxed{1.49 \times 10^{11} \, \text{m}} $

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Problem 4:

> You and your friend drove a total 32.0 km. You travel at 90.0 km/h. Your friend travels at 85 km/h. How long will your friend be waiting for you at the end of the trip?

#### Solution:

We need to find the time each person takes to travel 32.0 km.

- Your time:
$$
t = \frac{d}{v} = \frac{32.0}{90.0} = 0.3556 \, \text{h} \approx 21.33 \, \text{min}
$$

- Friend’s time:
$$
t = \frac{32.0}{85.0} = 0.3765 \, \text{h} \approx 22.59 \, \text{min}
$$

Wait time = Friend’s time – Your time:
$$
0.3765 - 0.3556 = 0.0209 \, \text{h} \approx 1.25 \, \text{min}
$$

But let’s check the answer format. It seems the student wrote:

> $ \frac{32.0 \, \text{km}}{90.0 \, \text{km/h}} = 0.3556 \, \text{h} $,
> $ \frac{32.0 \, \text{km}}{85.0 \, \text{km/h}} = 0.3765 \, \text{h} $,
> Difference = $ 0.0209 \, \text{h} \approx 1.25 \, \text{min} $

So the friend waits for you, meaning you arrive first.

Wait time = $ 0.3765 - 0.3556 = 0.0209 \, \text{h} $

Convert to seconds or minutes:
$$
0.0209 \, \text{h} \times 60 = 1.254 \, \text{min} \approx 1.25 \, \text{min}
$$

✔ Answer: $ \boxed{1.25 \, \text{min}} $ (or about 75 seconds)

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Problem 5:

> Ann is driving down a street at 55 km/h. Suddenly, a child runs into the street. If it takes Ann 0.75 seconds to react and brake, how many meters will the car travel before she begins to brake? (If the shoes skid at a rate of 1.5 m/s², how long will it take her to stop?)

#### Part a: Distance during reaction time

- Speed = 55 km/h → convert to m/s:
$$
55 \, \text{km/h} = \frac{55 \times 1000}{3600} = 15.28 \, \text{m/s}
$$

- Time = 0.75 s

$$
\text{Distance} = v \times t = 15.28 \times 0.75 = 11.46 \, \text{m}
$$

✔ Answer: $ \boxed{11.46 \, \text{m}} $

#### Part b: Time to stop after braking

- Initial speed $ v_0 = 15.28 \, \text{m/s} $
- Final speed $ v = 0 $
- Deceleration $ a = -1.5 \, \text{m/s}^2 $

Use:
$$
v = v_0 + at
\Rightarrow 0 = 15.28 - 1.5t
\Rightarrow t = \frac{15.28}{1.5} = 10.19 \, \text{s}
$$

✔ Answer: $ \boxed{10.19 \, \text{s}} $

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Problem 6:

> You drive your car for 2.0 h at 40.0 km/h, then for 2.0 h at 60.0 km/h. What is your average velocity?

#### Solution:

Average velocity = Total displacement / Total time

- First part: $ d_1 = 40.0 \, \text{km/h} \times 2.0 \, \text{h} = 80.0 \, \text{km} $
- Second part: $ d_2 = 60.0 \, \text{km/h} \times 2.0 \, \text{h} = 120.0 \, \text{km} $
- Total distance = $ 80 + 120 = 200 \, \text{km} $
- Total time = $ 2.0 + 2.0 = 4.0 \, \text{h} $

$$
\text{Average velocity} = \frac{200 \, \text{km}}{4.0 \, \text{h}} = 50.0 \, \text{km/h}
$$

✔ Answer: $ \boxed{50.0 \, \text{km/h}} $

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✔ Final Answers Summary:

1. $ -85 $ miles
2. a. $ 0 $ m; b. $ 6.0 $ m
3. $ 1.49 \times 10^{11} $ m
4. $ 1.25 $ min
5. a. $ 11.46 $ m; b. $ 10.19 $ s
6. $ 50.0 $ km/h

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