Finding the maximum velocity ended up being to involved (lots of physics) for th

Question

Finding the maximum velocity ended up being to involved (lots of physics) for this course. .

The path of our roller coaster (picking up in the first ascent and ending before the final coast into the loading station) will follow the function:

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The path of our roller coaster (picking up in the first ascent and ending before the final coast into the loading station) will follow the function: f(x) = 42x5 + 30x4 - 1060x3 - 204x2 + 4848x + -328

Explanation / Answer

to find maximum velocity,

firstly we need to find exp of velocity.

So, f'(x) gives us velocity.

f'(x) = 210x^4 + 120 x^3 - 3180x^2 - 408 x + 4848

Now, to find maximum velocity,

we need to find f"(x) and make it equal to 0.

f"(x) = 840x^3 + 360x^2 - 6360x - 408

equating it equal to 0,

we get

x = -2.944

x = -0.064

x = 2.579

Now to find which point is maxima, we need to find f"'(x),

f"'(x) = 120(21x^2 + 6x - 53)

Now we need to find values at x,

At x= -2.944

f"'(x) = 13361.5

f"'(x)>0 so, it is not maxima

At x= - 0.064

f"'(x) = -6395.76 means maxima

At x = 2.579,

f"'(x) =12 258. So, no maxima at this point.

Now we need to find value of velocity at x= - 0.064

Now, f'(x) = 4841.06

At x= -6,

f'(x) = 139056

At x= 4,

f'(x) = 13776

So, fi=unction has max velo at x= -6 and v = 139056

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