Solve 2D Motion Q: Wiley E Coyote & Roadrunner

In summary, Wiley E Coyote is attempting to catch the roadrunner using jet powered roller skates. The roadrunner is able to reach the cliff before the coyote by using a minimum speed calculated from the given parameters [ac, D, H]. The coyote will land at the bottom of the canyon after making a sudden turn at the cliff, with a velocity determined by the velocity at the end of the first part of its trajectory.
  • #1
oreosama
51
0

Homework Statement



wiley e coyote is out once more to try to capture the elusive roadrunner. the coyote wears a pair of acme jet powered roller skates, which provide a constant horizontal acceleration ac. the coyote starts off at rest D meters from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff. the cliff is H meters above the canyon floor. given [ac, D, H], determine:

if the roadrunner moves at constant speed, what should the roadrunners min speed be in order to reach the cliff before the coyote?

as usual the roadrunner is saved by making a sudden turn at the cliff. determine where the coyote lands in the canyon. (assume skates still on)

the velocity of the coyote just before hitting the canyon floor

Homework Equations



vx=v0x+at
x=x0+v0x*t+1/2*a*t^2
vx^2=v0x^2 + 2a(x - x0)

and y equivilents



The Attempt at a Solution




for a, i solved for time(t1) as t1= sqrt(2D/ac), which i then use for velocity of roadrunner(Vr) as Vr= D/sqrt(2D/ac)

for b & c, i used info from solving things in a as info to parameters:

from D to bottom of cliff

x0= D
x= D2
Vx0= acsqrt(2D/ac)
Vx= ?
a = ac

y0=0
y= H
Vy0 = 0
Vy = ?
a = g

t=t2

trying to use my y-based equations to solve for t2 results in them collapsing

Vy= 0 + gt2
Vy^2 = 0^2 + 2gH

g^2*(t2)^2 = 2gH

H = (g*t2^2)/2

(g*t2^2)/2 = 0 + 0 + 1/2*g*(t2)^2

everything cancels :<
 
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  • #2
The trajectory will have two distinct parts. Find the velocity at the end of the first part. The second part, due to the jets still on, will be accelerated both horizontally and vertically. Use the final velocity of part one to find what is required.
 

FAQ: Solve 2D Motion Q: Wiley E Coyote & Roadrunner

1. How does Wiley E Coyote always fail to catch the Roadrunner?

This is due to the laws of physics, specifically the concept of inertia. As the Roadrunner is constantly moving, it has a greater inertia than the Coyote who is trying to catch it. This means that the Roadrunner will continue to move at a constant speed and direction, making it difficult for the Coyote to catch up.

2. Can Wiley E Coyote ever successfully catch the Roadrunner?

In theory, yes. If the Coyote were to apply a force or acceleration greater than the Roadrunner's inertia, he could potentially catch it. However, this would require a significant amount of energy and precise timing, making it nearly impossible in reality.

3. How does gravity affect Wiley E Coyote's attempts to catch the Roadrunner?

Gravity plays a significant role in the Coyote's attempts to catch the Roadrunner. As the Coyote is constantly falling towards the ground due to the force of gravity, he has to exert more force and energy to move in any direction. This makes it difficult for him to keep up with the Roadrunner's constant movement.

4. Why does Wiley E Coyote always fall off cliffs and get hit by his own traps?

Again, this is due to the laws of physics. As the Coyote is constantly moving and attempting to catch the Roadrunner, he is unable to accurately calculate his movements and often ends up falling off cliffs or getting hit by his own traps. This can be attributed to the concept of momentum and the Coyote's lack of control over it.

5. Can the Roadrunner ever be caught by Wiley E Coyote's inventions?

In the fictional world of the cartoon, it is highly unlikely that the Roadrunner will ever be caught by any of the Coyote's inventions. However, in reality, it is possible for the Coyote's inventions to successfully catch the Roadrunner if they are designed and executed correctly. It all comes down to proper planning and execution of the physics behind the inventions.

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