A motorcyclist is driving up a ramp that has a constant incline relative to level ground. After he rides 1000 feet on the ramp, he is 100 feet off the ground. What is his speed relative to the ground if he drives at 80 miles per hour up the ramp?

Question

amistre64 · Answer

using a right triangle, we sort of relationship can we construct to play with?

amistre64 · Answer

*what sort of ...

anonymous · Answer

I think it has to do with the derivative

amistre64 · Answer

it does, but we need to relate the information to something more equationy,  like may pythag thrm

amistre64 · Answer

spped relative to the ground, is that his horizontal speed? or vertical spped?

anonymous · Answer

I already solved for these values

|dw:1413570352884:dw|

anonymous · Answer

The height is 100 btw (it got cut off in the drawing)

amistre64 · Answer

it shows on my screen, might be a zoom thing on yours

amistre64 · Answer

well speed is a linear factor just as length is ... so whatever direction is being requested has the same ratio

1000          80
----- = -------
 100        vertical

or the other leg for the horizontal

amistre64 · Answer

or by trig:

80 cos(theta) 
or
80 sin(theta) are just as good

anonymous · Answer

I don't really understand

amistre64 · Answer

|dw:1413570512747:dw|

its just similar triangles

anonymous · Answer

I'm not sure that is what the question is asking. I am pretty sure it has to do with the derivative because that is what we learned in class.

amistre64 · Answer

well it does have to do with the derivative, but it can be simplified to a rather basic algebra/trig problem

anonymous · Answer

Can you explain how to use the derivative to get the answer, so I can understand the question in terms with what we are learning in class?

amistre64 · Answer

assuming speed relative to the ground is horizontal, or in the +x direction

cos(a) = x/r

take implicit derivative to get

-sin(a) a' = (rx'-xr')/r^2

since the angle doesnt change, a'=0; and r' is given as 80, r=1000

0 = (1000x'-80x)/(1000)^2

you determined x is about 995 sooo

0 = (1000x'-80(995))/(1000)^2

solve for x'

amistre64 · Answer

80 (995) = 1000 x'

which of course is the same ratio of similar triangles:

1000/995 = 80/x'

anonymous · Answer

So x' would be 398/5?

amistre64 · Answer

if relative to the ground is in the +y direction

sin(a) = y/r  same process

anonymous · Answer

I just realized something. At the beginning, wouldn't we have to convert 80 mph to ft/s?

amistre64 · Answer

no

anonymous · Answer

Oh okay. Just making sure. Thank you for your help.

amistre64 · Answer

the angle of the ramp is not changing, so all linear forces/measures define similar triangles ....

amistre64 · Answer

good luck