The graph below shows the velocity f(t) of a runner during a certain time interval:

2 answers:

5 0

For this case, the first thing we must do is take into account the definition of the axes:
the x axis is time in seconds
the y axis is velocity in meters per secon

Then, we look for the cut point with the y axis, which is given by:
(0, 2)
This means that the initial speed is 2 m / s.

We now look for the cut point with the x axis, which is given by:
(8, 0)
This means that the speed is 0 m / s, so the runner stopped when the time is 8 seconds.

Answer:
The initial velocity of the runner was 2 m / s, and the runner stopped after 8 seconds.

jeka57 [31]3 years ago

3 0

Answer:

The initial velocity of the runner was 2 m/s, and the runner stopped after 8 seconds.

Step-by-step explanation:

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by the double angle identity for sine. Move everything to one side and factor out the cosine term.

$10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0$

Now the zero product property tells us that there are two cases where this is true,

$\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}$

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of $\dfrac\pi2$ , so $x=\dfrac{(2n+1)\pi}2$ where $n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}$

which occurs twice in the interval $[0,2\pi)$ for $x=\arcsin\dfrac3{10}$ and $x=\pi-\arcsin\dfrac3{10}$ . More generally, if you think of $x$ as a point on the unit circle, this occurs whenever $x$ also completes a full revolution about the origin. This means for any integer $n$ , the general solution in this case would be $x=\arcsin\dfrac3{10}+2n\pi$ and $x=\pi-\arcsin\dfrac3{10}+2n\pi$ .