Minimum Speed for Motorcyclist to Successfully Cross a Ravine

Click For Summary

Homework Help Overview

The problem involves a motorcyclist attempting to cross a ravine that is 2.0 meters higher than the launch point, using a ramp angled at 40 degrees and spanning a width of 10.0 meters. The objective is to determine the minimum speed required for the motorcyclist to successfully clear the ravine.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to relate the horizontal and vertical motions of the motorcyclist. There are attempts to express the unknowns in terms of each other and to substitute variables into the equations.

Discussion Status

Some participants have provided guidance on how to set up the equations, noting that two equations can be used to solve for the two unknowns. There is acknowledgment of the method being correct, but concerns are raised about the calculated speed being insufficient for crossing the ravine.

Contextual Notes

Participants are working under the constraints of the problem's parameters and are questioning the accuracy of their calculations. There is an emphasis on ensuring the motorcyclist achieves the necessary height and distance to avoid falling into the ravine.

ladymiresa
Messages
4
Reaction score
0

Homework Statement


A motorcyclist must cross a ravine. The far side of the ravine is 2.0 meters higher than the launch point, a ramp that makes an angle of 40◦ with the horizontal. If the ravine is 10.0 meters wide, what minimum speed v must the biker have when leaving the ramp to successfully cross the ravine? Take the launch point at the end of the ramp as the coordinate origin.

Homework Equations


yf=yi+(vi)t-.5g(t^2)

The Attempt at a Solution



Well, I know that Vxi=vicos40 and vyi=visin40.
Known: xi=0, yi=0, ti=0, xf=10, yf=2, theta=40, ay=-9.8
I want to use to equation above, but I can't because I know neither the Vi nor the T. I thought I could figure out the T by saying that the initial vy was zero, but that definitely doesn't seem right because then only gravity would be acting on the bike and he has to GAIN 2m in height. So I'm a little stuck. I'd really appreciate any tips to help me figure out this problem!
 
Physics news on Phys.org
You have two equations, one in the y direction and one in the x direction. There are two unknowns they share.
 
The only equation in the x direction I can think of is xf=xi+(vx)i*t, and here I still have two unknowns (xfi and t).
 
ladymiresa said:
The only equation in the x direction I can think of is xf=xi+(vx)i*t, and here I still have two unknowns (xfi and t).
No, the two unknowns there are vx and t, surely? And you can express vx in terms of v and theta. So as I posted, both unknowns appear in both equations. Two equations, two unknowns.
 
Ah, yes, I definitely meant Vx and t for the unknowns.

So, if I say that vx=vicos40 then I can rewrite the equation as xf=xi+(vicos40)t... which can be rearranged to say t=xf]-xi/(vicos40).

then I think I can substitute into the first equation:
yf=yi+(visin40)(xf-xi/vicos40)+.5(-9.8)(xf-xi/vicos40)^2. So my only unkown is vi! ...Now, if only I can get the math right haha..

Ok, I calculated that Vi=9.92m/s. I think this makes sense. Thanks for helping me, I never would have thought to substitute the equations on my own.
 
ladymiresa said:
Ah, yes, I definitely meant Vx and t for the unknowns.

So, if I say that vx=vicos40 then I can rewrite the equation as xf=xi+(vicos40)t... which can be rearranged to say t=xf]-xi/(vicos40).

then I think I can substitute into the first equation:
yf=yi+(visin40)(xf-xi/vicos40)+.5(-9.8)(xf-xi/vicos40)^2. So my only unkown is vi! ...Now, if only I can get the math right haha..

Ok, I calculated that Vi=9.92m/s. I think this makes sense. Thanks for helping me, I never would have thought to substitute the equations on my own.

9.92m/s isn't fast enough. You'll end up in the ravine at that speed! Maybe check the maths.
 
ladymiresa said:
Ah, yes, I definitely meant Vx and t for the unknowns.

So, if I say that vx=vicos40 then I can rewrite the equation as xf=xi+(vicos40)t... which can be rearranged to say t=xf]-xi/(vicos40).

then I think I can substitute into the first equation:
yf=yi+(visin40)(xf-xi/vicos40)+.5(-9.8)(xf-xi/vicos40)^2. So my only unkown is vi! ...Now, if only I can get the math right haha..

Ok, I calculated that Vi=9.92m/s. I think this makes sense. Thanks for helping me, I never would have thought to substitute the equations on my own.
I haven't checked your arithmetic, but good to see that you get the method.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
7K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
1
Views
5K