Let v(t) = 1/π + sin(3t) represent the velocity of an object moving on a line. On the interval [π/2, π], what is the velocity when the acceleration is 3?

Answer:

[tex]\displaystyle \frac{1}{\pi}\approx0.318[/tex]

Step-by-step explanation:

The velocity of an object moving on a line is given by the function:

[tex]\displaystyle v(t)=\frac{1}{\pi}+\sin(3t),\; \frac{\pi}{2}\leq x\leq\pi[/tex]

And we want to determine the velocity of the object when its acceleration is 3.

So, we will need to determine our acceleration function. Remember that acceleration is simply the derivative of speed. Therefore, our acceleration function a(t) will be:

[tex]\displaystyle a(t)=v^\prime(t)=\frac{d}{dt}[\frac{1}{\pi}+\sin(3t)][/tex]

Differentiate. We will use the chain rule on the second part. Hence, our acceleration function is:

[tex]\displaystyle a(t)=3\cos(3t)[/tex]

We want to find the velocity when the acceleration is 3.

So, let’s set a(t) equal to 3 and determine at what time t our function equals 3. Hence:

[tex]3=3\cos(3t)[/tex]

Divide both sides by 3:

[tex]\cos(3t)=1[/tex]

Take the inverse cosine of both sides.

Remember that cosine is 1 for every 2π. Hence:

[tex]3t=2\pi n[/tex]

Where n is an integer.

Divide both sides by 3. Therefore, our solutions are:

[tex]\displaystyle t=\frac{2\pi n}{3}[/tex]

However, our domain was π/2≤x≤π.

Hence, we can test values of n such that it is within our domain.

Testing for n=0, 1, and 2, we see that:

[tex]\displaystyle t=0, \, t=\frac{2\pi}{3},\, t=\frac{4\pi}{3}...[/tex]

However, the only solution that is within our domain is the second solution.

Hence, the acceleration is 3 when the t is 2π/3.

Therefore, our velocity at that time t is:

[tex]\begin{aligned}\displaystyle v(\frac{2\pi}{3})&=\frac{1}{\pi}+\sin(3(\frac{2\pi}{3}))\\ &=\frac{1}{\pi}+\sin(2\pi)\\&=\frac{1}{\pi}+0\\&=\frac{1}{\pi}\approx 0.318\end{aligned}[/tex]

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