contestada

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)

Respuesta :

facundo3141592

Answer: (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)

     

jane3455

Answer:

c

Step-by-step explanation: