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mima031 · May 14, 2012, 07:23 AM

Martin is on his trail bike, approaching a creek bed that is 7 m wide. A ramp with an incline of 10 degrees has been built for daring people who try to jump the creek. Martin is traveling at his bike's maximum speed,40 km/h. (a) Should Martin attempt the jump or enphatically hit the brakes? (b) What is the minimum speed the bike must have to make his jump? (Assume equal elevations on either side of the creek)

Stratmando · May 14, 2012, 07:35 AM

Curious what results are, but jumped many bikes, I would say between 25 and 30 MPH. This is not a simple Physics question, as it likely won't figure in the ability to increase the distance by lifting on the bike at take off.
I'm curious the correct answer.

ebaines · May 14, 2012, 10:24 AM

When an object is tossed in the air at initial velocity V the time it takes to reach the apex of its arc is V/g, and the time it takes to fall back to earth is another V/g. Hence the total time in the air is 2V/g. Here the initial vertical velocity of the bike is 40 Km./hr times sin(10 degrees. So find the time the bike is in the air, then multiply that time by the horiziontal speed of the bike to find how far horizontally the bike travels while in the air.

Stratmando - while 40 Km/Hr is about 25 MPH and so may see "fast enough," the incline is only 10 degrees so the bike doesn't get much lift at all. If it was more like 20 degrees he'd have a better chance.

Stratmando · May 14, 2012, 10:44 AM

ebaines, I don't think this can be solved yet?
An Important factor is how high is the ramp, 10 degrees is important, but the height of the ramp at the top is missing(which would also determine the length of the ramp)?
Thanks.

ebaines · May 14, 2012, 11:26 AM

Stratmando - it is assumed that the take-off point and landing point are at the same elevations. Yes, it can be solved using basic physics concepts (as long as we can ignore wind resistance).

Stratmando · May 14, 2012, 06:25 PM

I got it.
Oh, The thing about this problem is when you are actually attempting this jump is, if you are feeling you are going slow, you can pull yourself and Bike/Motorcycle straight up at the end of the ramp, and get increased distance that I am not sure it could be figured.

He also Mentions hitting the brakes, When you are Jumping a Motorcycle at any good distance, your throttle AND Brakes can affect your Attitude, Land Real wheel wheel first/Smoothly.
If you give a little throttle in the air(the front raises up/rear end lowers), and if you tap the brakes the front will come down.

The Rear wheel want to stay the same speed, but when touching brakes or throttle, it adjust your attitude, it will definitely help you.

actuallyimcold · May 15, 2012, 01:31 AM

I got 50 km/h to get the bike to safe landing.

ebaines · May 15, 2012, 06:16 AM

Originally Posted by actuallyimcold

I got 50 km/h to get the bike to safe landing.

That's about right.

Stratmando · May 15, 2012, 10:57 AM

50 KPH equals about 31 MPH.
Could still be done at 30, you could make up speed by pulling up.
I didn't assume identical ramps as in Mountain Biking, Most Jumps, landing height is usually a lower height(not always).
I am most trying to say, The Input a rider can make, can throw a formula off.

spade18 · May 15, 2012, 12:46 PM

ebaines: Can you show the computation on how you solve this problem? :D

ebaines · May 15, 2012, 01:21 PM

Originally Posted by spade18

Can you show the computation

As pointed out in post #3 the time in the air is

$ t = 2 \frac {v \sin \theta} {g} $

and the horizontal distance traveled during that time is

$ d = vt \cos \theta $

Combine these to get:

$ d = \frac {2v^2 \cos \theta sin \theta} g $

Now rearrange to solve for v.

actuallyimcold · May 16, 2012, 06:31 AM

Originally Posted by ebaines

As pointed out in post #3 the time in the air is

$ t = 2 \frac {v \sin \theta} {g} $

and the horizontal distance traveled during that time is

$ d = vt \cos \theta $

Combine these to get:

$ d = \frac {2v^2 \cos \theta sin \theta} g $

Now rearrange to solve for v.

i used your formula and i got 0.39 s. It seemed unrealistic.

ebaines · May 16, 2012, 06:53 AM

Originally Posted by actuallyimcold

i used your formula and i got 0.39 s. It seemed unrealistic.

Yes - that's the time the bike is in the air. Remember that the ramp is only inclined at ten degrees, which is not very much at all. If you calculate how high the bike goes up in the air - at its highest point it's less than 0.2m off the ground!

actuallyimcold · May 16, 2012, 06:57 AM

OK. Another question, where does you're the first formula derived from? :)

ebaines · May 16, 2012, 07:17 AM

Originally Posted by actuallyimcold

ok. another question, where does your the first formula derived from? :)

$ \Delta v = at $

For an object launched vertically at speed $v_1$ , at the top of its arc its vertical locity is zero so $\Delta v = -v_1$ , and the acceleration is equal to -g. The time to reach the highest point is therefore

$ t_1 = \frac {v_1} g $

After reaching the apex the object then starts to fall back to the ground, and the time it takes to do that is the same as the above. So total time in the air is

$ t =2t_1 = 2 \frac {v_1} g $

actuallyimcold · May 16, 2012, 07:20 AM

Originally Posted by ebaines

$ \Delta v = at $

For an object launched vertically at speed $v_1$ , at the top of its arc its vertical locity is zero so $\Delta v = -v_1$ , and the acceleration is equal to -g. The time to reach the highest point is therefore

$ t_1 = \frac {v_1} g $

After reaching the apex the object then starts to fall back to the ground, and the time it takes to do that is the same as the above. So total time in the air is

$ t =t_1 + t_2 = 2 \frac {v_1} g $

now that's clearer. Thank you very much!