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2 years ago

Henry ran a 100 m race. Use the graph to answer the following:

Physics

1 answer:

horrorfan [7]2 years ago

6 0

Using the graph, which describes how Henry ran the 100m race;

a) It takes Henry 20seconds to run 100m

b) Henry's average speed over the race is; 5m/s.

According to the linear graph which describes the distance ran by Henry during the 100m race as a function of time.

a) Since the distance from start ran by Henry is plotted on the vertical axis, and the time is plotted on the horizontal axis;

To determine how long it took Henry to run 100m; The point corresponding to 100m is traced downward from the line of the graph and we find out that;

It takes Henry 20seconds to run 100m

b) Henry's average speed over the race is simply;

The slope of the distance-time graph.

Therefore,

Average speed = (100-0)/(20-0)

Average speed = 100/20

Average speed = 5m/s.

Therefore, Henry's average speed over the race is; 5m/s.

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3 years ago

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3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

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

Explanation:

The average velocity is total displacement divided by time:

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

And in the case of vertical $v_{avg}$

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

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:

$y=H+v_0t-\dfrac{1}{2} gt^2$

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,