Find the distance traveled in 45 seconds by an object traveling at a velocity of v(t) = 20 + 9cos(t) feet per second.

Take a look at your text. There must be a problem similar to this.

Consider an object traveling at constant velocity v = k. Then, each second, it travels k feet, so after t seconds, the distance traveled is k*t.

Now, suppose its speed increases by 1 ft/sec each second. So, a ballpark estimate for distance traveled after t seconds is
1 + 2 + 3 + 4 + ... + t = t(t+1)/2, or approximately 1/2 t^2.

Do you see a pattern forming here? Given a velocity function v, the distance s is figured by taking the integral. In this case,

v = 20 + 9 cos(t)
s = 20t + 9 sin(t) from 0 to 45 = s(45)-s(0)

s(0) = 0 so, the distance is just s(45) = 900 + 9sin(45) = 900 + 9*0.85 = 907.65 feet.

Well, if we want to find the distance traveled in 45 seconds, we need to integrate the velocity function over that time period. But before we do that, let me tell you a joke to lighten the mood.

Why don't scientists trust atoms?

Because they make up everything!

Alright, now let's get back to the problem. To find the distance traveled, we integrate the absolute value of the velocity function over the given time interval. Let's set up the integral:

∫[0 to 45] |20 + 9cos(t)| dt

Now, let's solve the integral and find out how far our intrepid object has traveled in those 45 seconds.

To find the distance traveled in 45 seconds, we need to integrate the velocity function over the given time interval.

Given the velocity function v(t) = 20 + 9cos(t) feet per second, we can integrate it over the interval [0, 45] to find the displacement.

To integrate the velocity function v(t), we'll integrate each term separately:

∫(20 + 9cos(t)) dt

The integral of the constant term 20 is 20t.

The integral of cos(t) is sin(t).

Therefore, the integral of v(t) = 20 + 9cos(t) is:

= 20t + 9sin(t)

Now, we'll evaluate this expression at the endpoints of the interval [0, 45]:

Displacement at t = 45 seconds - Displacement at t = 0 seconds

= (20(45) + 9sin(45)) - (20(0) + 9sin(0))

= (900 + 9√2) - 0

= 900 + 9√2

So, the object travels a distance of 900 + 9√2 feet in 45 seconds.

To find the distance traveled in 45 seconds, we need to integrate the velocity function with respect to time over the given time interval.

The velocity function is v(t) = 20 + 9cos(t) feet per second.

To integrate this function, we first find the antiderivative of the function.

The antiderivative of 20 with respect to t is 20t.
The antiderivative of 9cos(t) with respect to t is 9sin(t).

So, the antiderivative of v(t) = 20 + 9cos(t) is s(t) = 20t + 9sin(t) + C, where C is the constant of integration.

To find the constant C, we need an initial condition. Since we don't have that information, we leave it as C for now.

Now, we can find the distance traveled by evaluating the antiderivative function at the endpoints of the time interval and taking the difference.

Let's evaluate the antiderivative function at t = 45 seconds:

s(45) = 20(45) + 9sin(45) + C

To find the value of sin(45), we need to convert 45 degrees to radians. Since 180 degrees = pi radians, we have:

45 degrees = (45*pi)/180 radians = pi/4 radians.

So, sin(45) = sin(pi/4) = 1/sqrt(2).

Now, let's substitute the values into the equation:

s(45) = 20(45) + 9(1/sqrt(2)) + C

We still have the constant C in the equation, which we cannot determine without more information.

Therefore, the distance traveled in 45 seconds by an object with a velocity function v(t) = 20 + 9cos(t) feet per second, is given by the expression s(45) = 20(45) + 9(1/sqrt(2)) + C.