Question

005 10.0 points
a reconnaissance plane flies 570 km away from its base at 748 m/s, then flies back to its base at 1122 m/s.
what is its average speed?
answer in units of m/s.
006 10.0 points
the graph shows position as a function of time for two trains running on parallel tracks. at time t = 0 the position of both trains is 0 (at the origin).
which is true?

1. somewhere before time t_b, both trains have the same acceleration.
2. both trains speed up all the time.
3. both trains have the same velocity at some time before t_b.
4. in the time interval from t = 0 to t = t_b, train b covers more distance than train a.
5. at time t_b, both trains have the same velocity.

007 (part 1 of 6) 10.0 points
the angle θ is given in the figure below.
008 (part 2 of 6) 10.0 points
a) find cos θ.
b) find sin θ.
009 (part 3 of 6) 10.0 points
c) find tan θ.
010 (part 4 of 6) 10.0 points
d) find sec θ.
011 (part 5 of 6) 10.0 points
e) find csc θ.
012 (part 6 of 6) 10.0 points
f) find cot θ.

Explanation:

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Step1: Calculate total distance

The plane flies 570 km away and then 570 km back, so the total distance $d = 570+570=1140$ km.

Step2: Calculate time for each - part of the journey

The time taken to fly away $t_1=\frac{570\times1000}{748}$ s (converting km to m), and the time taken to fly back $t_2=\frac{570\times1000}{1122}$ s.

Step3: Calculate total time

$t = t_1 + t_2=\frac{570\times1000}{748}+\frac{570\times1000}{1122}$ s.

Step4: Calculate average speed

The average - speed formula is $v=\frac{d}{t}$. First, simplify $t=\ 570\times1000\times(\frac{1}{748}+\frac{1}{1122})=570\times1000\times\frac{1122 + 748}{748\times1122}=570\times1000\times\frac{1870}{748\times1122}$. Then $v=\frac{1140\times1000}{570\times1000\times\frac{1870}{748\times1122}}=\frac{2\times748\times1122}{1870}\approx902.4$ m/s.

The slope of a position - time graph represents velocity. At the intersection of the two graphs (at time $t_B$), the slopes of the two lines are different, so the velocities are different. The acceleration is the second - derivative of the position with respect to time, and we cannot say that they have the same acceleration before $t_B$. Train A has a steeper slope (higher velocity) in the later part of the time interval, so it covers more distance from $t = 0$ to $t=t_B$. The correct statement is that both trains have the same velocity at some time before $t_B$ because the slopes of the two curves are equal at the point of intersection of their tangents at some time before $t_B$.

Step1: Identify adjacent and hypotenuse

In a right - triangle, if the adjacent side to the angle $\theta$ is $x = 8$ and the hypotenuse is $r=\sqrt{8^{2}+2^{2}}=\sqrt{64 + 4}=\sqrt{68}=2\sqrt{17}$ according to the Pythagorean theorem $r=\sqrt{x^{2}+y^{2}}$.

Step2: Calculate cosine

The cosine of an angle in a right - triangle is given by $\cos\theta=\frac{x}{r}$. So $\cos\theta=\frac{8}{2\sqrt{17}}=\frac{4}{\sqrt{17}}=\frac{4\sqrt{17}}{17}$.

Answer:

902.4 m/s

006