Projectile Motion of roller skates

In summary: The final velocity is 45.83 m/s; you can break this into x and y components, and use the latter to find the time in the air. Then use the former to find the total time. Then use this total time to find the distance traveled in the x direction. In summary, the determined Wile E. Coyote is attempting to catch the elusive roadrunner using new Acme power roller skates with a constant horizontal acceleration of 15 m/s2. The coyote starts at rest 70 m from the edge of a cliff and the roadrunner moves with constant speed. To reach the cliff before the coyote, the roadrunner must have a minimum speed. Additionally, if the cliff is
  • #1
LastBloomingFlower
15
0
4) The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of Acme power roller skates, which provide a constant horizontal acceleration of 15 m/s2, as shown in Figure P3.73. The coyote starts off at rest 70 m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff.

(a) If the roadrunner moves with constant speed, find the minimum speed the roadrunner must have in order to reach the cliff before the coyote.

b) If the cliff is 100 m above the base of a canyon, find where the coyote lands in the canyon. (Assume that his skates are still in operation when he is in "flight" and that his horizontal component of acceleration remains constant at 15 m/s2.)

I don't understand even how to begin it...
 
Physics news on Phys.org
  • #2
Begin by writing the equations of motion for each animal. Which one starts from rest? Which one has acceleration? What happens in the vertical direction to Mr. Coyote?
 
  • #3
LastBloomingFlower said:
4) The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of Acme power roller skates, which provide a constant horizontal acceleration of 15 m/s2, as shown in Figure P3.73. The coyote starts off at rest 70 m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff.

(a) If the roadrunner moves with constant speed, find the minimum speed the roadrunner must have in order to reach the cliff before the coyote.

b) If the cliff is 100 m above the base of a canyon, find where the coyote lands in the canyon. (Assume that his skates are still in operation when he is in "flight" and that his horizontal component of acceleration remains constant at 15 m/s2.)

I don't understand even how to begin it...
Part A works in only one dimension (let's call it x). The roadrunner has constant non-zero velocity (zero acceleration), which implies an initial non-zero velocity. Coyote has an initial velocity of zero, but uniform acceleration. Find equations that will plot out exactly what it will take for both Coyote and Roadrunner to have the same position at the lip of the cliff.

Part B involves only Coyote. Break his velocity into y- and x- components. Use the former to determine how long it will take before he hits the ground, and then plug this time into an equation along with the latter to solve for his distance traveled.
 
  • #4
Well I have for the coyote
t=3.06s
A= 15 m/s/s
Vf=45.9
Vi=0
d=70

Roadrunner
t=3.06
a=0
Vi=?
Vf= ?
d=70m

Am i doing it right so far?
 
  • #5
LastBloomingFlower said:
Well I have for the coyote
t=3.06s
A= 15 m/s/s
Vf=45.9
Vi=0
d=70

Roadrunner
t=3.06
a=0
Vi=?
Vf= ?
d=70m

Am i doing it right so far?
That all seems reasonable so far. Make it a habit to include units everywhere (e.g. on your final velocity, displacement, and time above).
 
  • #6
Geoffjb, have you worked out this problem at all? Just wondering. :)
 
  • #7
For the second part I found the time to be 1.71s and Vf to be 71.42 m/s and Vi to be 45.83m/s all of this beinbg in the y-component. So the x-component would have t=1.17s and A= 0 but how would I find d or any velocities?
 
  • #8
LastBloomingFlower said:
For the second part I found the time to be 1.71s and Vf to be 71.42 m/s and Vi to be 45.83m/s all of this beinbg in the y-component. So the x-component would have t=1.17s and A= 0 but how would I find d or any velocities?
The problem says you should assume the horizontal acceleration remains constant for the whole time. You could do the problem either way, but I assume you want the answer to the problem as given. You know how long it took to reach the edge of the cliff, and you know the time in the air. The total time is the sum of the two. The same equation that worked for the x motion on top of the cliff keeps working the whole time.
 

What is projectile motion?

Projectile motion is the motion of an object that is thrown or launched into the air and then moves under the influence of gravity, without any additional propulsion.

How does projectile motion relate to roller skates?

When a person wearing roller skates pushes off the ground, they become a projectile and move through the air under the force of gravity. The motion of the roller skates can be described using the principles of projectile motion.

What factors affect the projectile motion of roller skates?

The factors that affect the projectile motion of roller skates include the initial velocity, angle of launch, and gravity. Other factors such as air resistance and surface friction can also play a role.

How can we calculate the trajectory of roller skates in projectile motion?

The trajectory of roller skates in projectile motion can be calculated using equations from physics, such as the equations for displacement, velocity, and acceleration. These equations take into account the initial conditions and the forces acting on the roller skates.

What are some applications of understanding projectile motion of roller skates?

Understanding projectile motion of roller skates can be useful in various fields such as sports, engineering, and physics. For example, it can help in designing roller coasters, analyzing the trajectory of a thrown ball in sports, or predicting the motion of objects in space.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
845
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top