give me your thoughts about this: Chapter 2: Problem 6

According to a rule-of-thumb, each five seconds between a lightning flash and the following thunder gives the distance to the flash in miles. (a) Assuming that the flash of light arrives in essentially no time at all, estimate the speed of sound in m/s from this rule. (b) What would be the rule for kilometers?

Solution:

In the given problem, we are told that the speed of light according to the rule-of-thumb is 1 mile per 5 seconds. We are asked to find the speed of sound when the flash of light arrives in no time at all in m/s. After finding the speed we are asked to tell about the rule for kilometers.

Firstly, lets find the speed of light. The values for distance and speed are given in the problem. We can use the following formula:

Speed = Distance/Time

As the distance and time given in the question are 1 mile and 5 seconds, we can substitute the values in the formula:

Speed = 1mi/5sec

Speed = 0.2mi/s

As we are asked to find the speed of sound in m/s units. So, here I multiplied the result with 1610:

Speed=0.2*1610m/s

Speed = 322m/s

Now, we have found the speed of sound. The speed implies that the sound travels a distance of 322m in 1 sec. So, it would travel 966m in 3s. Hence, we can conclude the rule for kilometer to be approximately 1km every 3 s.

2) Chapter 2: Problem 12

Calculate the average speed and average velocity of a complete round trip in which the outgoing 250 km is covered at 95 km/h, followed by a 1.0-h lunch break, and the return 250 km is covered at 55 km/h.

Solution:

In this problem, we are given that a complete round trip is made. The outgoing and returning distance are the same that is 250 km each way.

The total distance traveled is (250 km outgoing + 250 km returning) 500km.

The velocity in outgoing trip is (v₁) 95 km/h and that in returning trip is (v₂) 55 km/h. As the starting and ending points of the trip are the same because it made a round trip, so the displacement is 0. We are asked to find average speed and average velocity.

We can use the following formulas:

Average Speed = Total distance/ Total time

Average Velocity = Displacement/ Change in time

The time is not provided in the problem, so lets find the time first and then proceed with the formulas.

During outgoing portion,

v₁ = x₁/t₁

t₁ = x₁/v₁

t₁ = 250 km/ 95 km h^-1

t₁ = 2.632 h

During returning portion,

v₂ = x₂/t₂

t₂ = x₂/v₂

t₂ = 250 km/ 55 km h^-1

t₂ = 4.545h

The total time including the lunch break is:

t_total = t₁ + t_lunch + t₂

t_total = (2.632 + 1.0 +4.545)h = 8.177 h

Using the time we just found and the information from the problem, we can substitute the values in the formulas:

Average speed = 500km/8.177h = 61 km h^-1

Average velocity = 0km/8.177h = 0

3) Chapter 3: Problem 72

A rock is kicked horizontally at 15 m/s from a hill with a 45 slope (Fig. 3-58). How long does it take for the rock to hit the ground?

Solution:

In this problem, we are provided with the information that the rock is kicked with a velocity of 15 m/s from a hill whose slope is 45. We need to find the time the rock takes to hit the ground.

Let's take the launch point be the origin of coordinates, with right and upward as the positive directions.

Now, let's define equations of line representing ground and motion of the rock.

Equation of line representing ground :

y _ground = -x

Equations of line representing motion of rock:

y_rock = v_o t and y_rock =-1/2 gt²

The equations of line representing motion of rock can be combined to get:

y_rock = -1/2g/v_o ²* x²_rock

By equating the two equations for y, we get:

y_rock = y _ground

-1/2g/v_o ²* x²_rock= -x

x= 2v_o ²/g

t = x/v_o= 2v_o /g = 2(15 m/s)/9.80 m/s²= 3.1 s

So, it take 3.1 s for the rock to hit the ground.