Beth leaves Muskegon, 30 mile north of Holland, traveling at 60 mph.

Alvin leaves Holland traveling north at V=20t+40 mi/hr.

When will Alvin pass Beth?
How far from Holland will they be?

:I think Beth's distance is d=60t-30 (using Holland as the frame of reference). Should I just take the anti-derivative of Alvin's velocity and then set those functions equal to each other and solve for time? Thanks.

No derivatives or Calculus needed here, it is a basic algebra problem.

I assume they left at the same time,
so they travelled the same time when they pass

distance covered by Beth = 60t
distance covered by Alving = 20t+40

that difference is 30 miles
20t + 40 - 60t = 30
-40t = -10
t = 1/4

They pass each other after 15 minutes
so distance that Beth went is (1/4)(60) = 15 miles

So they pass each other 15 miles north of Holland

you are wrong.

the answer is 3 hours and was achieved by taking the anti-derivative like I said.

Beth's distance is 60t+30.
To get Alvin's distance you must take the anti-derivative (10t^2 +40t +C) C=0.

Then set the equations equal to each other and solve for t. t= 3, -1

Your answer did not make sense. They pass each other 210 mile away from Holland.

Yes, you are on the right track. To find when Alvin will pass Beth, you need to set their distances equal to each other and solve for time. Here's how you can do it step-by-step:

Step 1: Set up the distance equation for Beth:
The distance traveled by Beth from Holland can be represented as:
d(beth) = 60t - 30 miles

Step 2: Set up the distance equation for Alvin:
The distance traveled by Alvin from Holland can be represented as:
d(alvin) = (20t + 40) miles

Step 3: Set the two distance equations equal to each other:
Set d(beth) = d(alvin):
60t - 30 = 20t + 40

Step 4: Solve for time:
Combine like terms on one side:
60t - 20t = 40 + 30
40t = 70

Divide both sides by 40:
t = 70/40
t = 7/4 or 1.75 hours

Step 5: Calculate the distance from Holland:
Substitute the value of t back into either Beth's or Alvin's distance equation to find the distance at that time:
Using Beth's equation:
d(beth) = 60(1.75) - 30
d(beth) = 105 - 30
d(beth) = 75 miles

So, Alvin will pass Beth after 1.75 hours and they will be approximately 75 miles from Holland.

Yes, you are on the right track with setting up the equations based on their distances from Holland as the frame of reference. To solve for when Alvin will pass Beth, you need to equate their distances and solve for time.

First, let's start with Beth's equation:
d = 60t - 30

Now, let's find Alvin's equation by integrating his velocity function:
v = 20t + 40
To integrate this, you can use the power rule for integration:
∫(20t + 40) dt = 10t^2 + 40t + C

Since we are only interested in the position, we'll ignore the constant C. So Alvin's equation becomes:
d = 10t^2 + 40t

Now, let's set the two equations equal to each other and solve for time:
60t - 30 = 10t^2 + 40t

To solve this quadratic equation, let's rearrange it to standard form:
10t^2 + 40t - 60t + 30 = 0
10t^2 - 20t + 30 = 0

Unfortunately, this quadratic equation can't be factored easily, so we have to use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 10, b = -20, and c = 30. Plugging these values into the quadratic formula, we get:
t = (-(-20) ± √((-20)^2 - 4 * 10 * 30)) / (2 * 10)
t = (20 ± √(400 - 1200)) / 20
t = (20 ± √(-800)) / 20

Notice that we have a negative value inside the square root, which means there are no real solutions to this equation. This suggests that Alvin will never pass Beth.

Therefore, they will not meet each other, and the distance between them will keep increasing as time goes on.