The position (feet traveled) of a car is given by the equation s(t)=(1/4)t^2+1

find the time when the car is going the same speed as its average speed over the interval 0 to 10 seconds
t=0
t=2.5
t=5
t=10
Never

I think it's either t=5 or t=2.5?
The end points are (0,1) and (10,26) so the average speed would be 2.5 right? The derivative is x/2 so would the average speed just be 2.5 or would you use the derivative (5/2=2.5) and then the answer would be 5?

the speed according to your function is t/2, not x/2

so you want
t/2 = 2.5
t = 5 <------ that is the time when the two speeds are the same
and the instantaneous speed at that time is t/2 = 5/2 or 2.5 ft/s

But it asked for the time when they were equal and that is 5 seconds

Square millimeters per millimeter

Sorry, to make it easier in my work I changed t to x. Thank you so much for your help!

Also, what could be the units for dy/dx if y is the surface area of a tumor and x is the radius of the tumor? Would it be meters per radian?

(Square millimeters per millimeter
Millimeters per centimeter
Meters per second
Gallons per hour
Or meters per radian)

I'm not sure about the exact answer, but I can certainly offer a humorous perspective on the situation!

Imagine the car trying to catch up to its own average speed over the interval. It's like a cat chasing its own tail, but in this case, it's a car chasing its own speed! If the car reached the same speed as its average, it would be a truly amusing sight to behold.

In terms of the actual solution, I'll leave that to you and your mathematical prowess. But remember, even if the car never reaches the same speed as its average, it can always become a famous racecar driver in the world of comedy racing!

To determine the time when the car is going at the same speed as its average speed over the interval 0 to 10 seconds, we need to find the derivative of the position function and then compare it to the average speed.

The position function is given as s(t) = (1/4)t^2 + 1.

Let's find the derivative of s(t):
s'(t) = d/dt [(1/4)t^2 + 1]
= (1/4) * d/dt(t^2) + d/dt(1)
= (1/4) * 2t + 0
= (1/2)t

Now, we need to calculate the average speed over the interval 0 to 10 seconds. The average speed is given by the change in position divided by the change in time:
Average Speed = (s(10) - s(0)) / (10 - 0)

Substituting the values into the equation, we have:
Average Speed = (26 - 1) / (10 - 0)
= 25 / 10
= 2.5

Therefore, the average speed over the interval is 2.5 ft/s.

To find the time when the car is going at the same speed as its average speed, we need to set the derivative equal to the average speed and solve for t:
(1/2)t = 2.5

Dividing both sides by (1/2), we get:
t = 5

So, the correct answer is t = 5. At t = 5 seconds, the car is going at the same speed as its average speed over the interval 0 to 10 seconds.