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Scorpion4ik [409]

2 years ago

7

You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 km/h (65mi/h), and the trip takes

2 hr and 20 min. On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 km/h (50 mi/h). How much longer does the trip take?

2 answers:

likoan [24]2 years ago

7 0

Answer:

Given the second case mentioned, the trip will take:

<u>1 hour and 10 minutes more than in the first case, that is, it will take 3 hours with 30 minutes in total.</u>

Explanation:

To calculate how much longer someone will take on that journey, you must calculate the distance traveled, using the aforementioned initial speed, however, the time must be converted into a number that you can calculate, this is done using a rule of three:

60 minutes = 1
20 minutes = X

X = 20 minutes / 60 minutes = 0.33333

Now it is calculated:

Kilometers traveled = 105 km / h (speed used) * 2,333 hours (the change from 2 hours and twenty minutes to a number) = 245 km

Now you just need to divide the total distance by the new speed to find the total time spent now:

Total time case 2 = 245 Km / 70 Km / h = <u>3.5 hours</u> (which is equal to <u>3 hours and 30 minutes</u>)

Since the time of the first case was 2 hours 20 minutes, the difference is <u>1 hour 10 minutes</u>.

Paha777 [63]2 years ago

5 0

Answer:

3 hours and 28 minutes.

Explanation:

Average speed is gu=iven by,

Average speed = $\frac{Total distance}{Total time}$

∴ Total distance = Average speed × Total time

Average speed = 105 $\frac{km}{h}$

Total time = 2 hour and 20 minutes = 2.33 hour

Total distance = 105 × 2.33

Total distance = 244.65 km

The total distance between San Diego and Los Angeles is 244.65 km.

Due to heavy traffic the average speed is decreased to 70 $\frac{km}{h}$ .

Average speed = $\frac{Total distance}{Total time}$

∴ Total time = $\frac{Total distance}{Average speed }$

Total time = $\frac{244.65}{70}$

Total time = 3.495 hours

Total time = 3 hours and 28 minutes

Thus due to heavy traffic the time taken to travel from between San Diego and Los Angeles is 3 hours and 28 minutes.

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$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

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The average velocity is total displacement divided by time:

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$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

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The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

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