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s2008m [1.1K]
3 years ago
5

A car moves at a speed of 40 miles per hour for half an hour and then at 60 miles per hour for two hours. What is the average sp

eed, measured in miles per hour, for the entire trip?
Physics
1 answer:
katovenus [111]3 years ago
7 0

Answer:

The average speed for the entire trip is 56 miles/hour.

Explanation:

It is given that,

Initially, the car moves with a speed of, v₁ = 40 miles/hour

Initial time, t₁ = 0.5 hours

Distance covered during time t₁,

d_1=v_1t_1

d_1=40\times 0.5

d_1=20\ miles

Then it travels with a speed of, v₂ = 60 miles/hour

Then time taken, t₂ = 2 hours

Distance covered during time t₂,

d_2=v_2t_2

d_2=60\times 2

d_2=120\ miles

The average speed of the car can be calculated as :

v=\dfrac{d_1+d_2}{t_1+t_2}

v=\dfrac{120+20}{0.5+2}

v = 56 miles/hour

So, the average speed of the car for the entire trip is 56 miles/hour. Hence, this is the required solution.  

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A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
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Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

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