The acceleration y, in meters per second squared, of an object after x seconds is given by y = 7 sin (pi/4 x). During the first 10 seconds, over which intervals is the acceleration increasing? (2, 6)
(4, 8)
(0, 2) and (6, 10)
(0, 4) and (8, 10)

(0, 2) and (6, 10)

Step-by-step explanation:

The acceleration is:

y = 7*sin( x*pi/4).

Over which interval is the acceleration increasing?

First, let's study the sin function:

Sin(θ) increases between -pi/2 and pi/2

Sin(θ) decreases between pi/2 and (3/2)*pi.

And so on.

Here we have:

θ = x*(pi/2)

When x = 0, θ = 0.

And sin(θ) is increasing in θ = 0.

Then we must choose one of the options that start with x = 0.

now it will stop increasing when:

θ = pi/2 = x*(pi/4)

     x = 2.

So the acceleration is increasing in the segment (0,2)

And it will start increase again when:

θ = (3/2)*pi = x*(pi/4)

     (3/2)*pi*(4/pi) = 6 = x.

So at x= 6 the acceleration starts increasing again, and the acceleration is only deffined until x = 10, then

the correct option is: (0, 2) and (6, 10)