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iris [78.8K]
3 years ago
12

Brad walks and jogs to schooll every day. He averages 5 km/hr walking and 9 km/hr jogging. The distane from home to shool is 6 k

m, and the total trip takes 1 hour. How far does Brad walk?
Physics
1 answer:
Vitek1552 [10]3 years ago
5 0
We know that
Distance = speed x time
Let w be the time Brad spent walking. The time spent jogging will be 1 - w
6 = 5w + 9(1 - w)
w = 0.75 hours
Distance walked = 0.75 x 5
= 3.75 km
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Answer:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

\int\limits^a_bγhxdy

we know that γ=62.5 lb/ft^{3}

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

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x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}

when substituting the x and y-values given on the graph, we get that the slope is:

m=\frac{0-6}{7-0}=-\frac{6}{7}

once we got this slope, we can substitute it in the point-slope form of the equation:

y_{2}-y_{1}=m(x_{2}-x_{1})

which yields:

y-6=-\frac{6}{7}(x-0)

which simplifies to:

y-6=-\frac{6}{7}x

we can now solve this equation for x, so we get that:

x=-\frac{7}{6}y+7

with this last equation, we can substitute everything into our integral, so it will now look like this:

\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy

Now that it's all written in terms of y we can now simplify it, so we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}

By using the fundamental theorem of calculus we get:

62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]

When solving we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

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