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Motion with constant acceleration problem

  • Thread starter flanders
  • Start date
  • #1
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Homework Statement


Hi. I've had some trouble solving this exercise, and I simply can't find the right solution. Hopefully you might help me out...?

Here's the task:

A plane (not airplane) has an inclination that allows a particle sliding on the slippery slope, and has an acceleration of 4.00 m / s ^ 2 The sloping surface is 1.28 m long.
The task:

Particle (V01) will be sent straight up the inclined plane with an initial speed of 5.00 m /s. Another particle is released at the same time without starting speed from the top of the inclined plane. See figure (Particle 1 is the red dot, particle 2 is the green one)

1. Where will the particles collide?
2. What kind of speed must the lower particle have if both particles were to collided in the middle of the slippery slope?



Homework Equations


X = X0 + V0t + (1/2)at^2
V = V0 + at



The Attempt at a Solution



In my mind, I think that both time and where they collide must be the same. Therefore X(particle1) = X(particle 2)

I've tried to solve it like this, particle 1 to the left, particle 2 to the right:

X0 + Vot + (1/2)at^2 = X0 + Vot + (1/2)at^2

This will leave me with a time, lets call it T. To make sure that it's the correct answer, I've tried to put T into both equations separately, to check if they collide at the same X. As you've probably understood, they don't.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

Last edited:

Answers and Replies

  • #2
gneill
Mentor
20,781
2,759
I've tried to solve it like this, particle 1 to the left, particle 2 to the right:

X0 + Vot + (1/2)at^2 = X0 + Vot + (1/2)at^2
But the particles are not starting with the same Vo and X0. You need to put your given data for each particle into the appropriate places to form the required equation.
 
  • #3
10
0
Okay, here is my solution:

Particle 1:
a = -4
v0 = 5
x0 = 0

Particle 2:
a = 4
v0 = 0
x0 = 1,28

Therefore;

Particle 1 to the left, particle 2 is on the right:


0 + 5t + 0,5*(-4)t^2 = 1,28 + 0t + 0,5*4t^2


5t - 2t^2 = 2t + 1,28


-4t^2 + 5t - 1,28 = 0

This leaves me with t = 0,36 or t = 0,89.

If I then put the t into X=X0 + V0t + 0,5at^2 to find the X, the result seems to be out of range....

Help??? :)
 
  • #4
gneill
Mentor
20,781
2,759
Particle 1:
a = -4 ---> acceleration is leftwards. Okay.
v0 = 5 ---> inital velocity is rightwards. Okay.
x0 = 0 ---> starting at origin. Okay.

Particle 2:
a = 4 ---> acceleration is rightwards! Oops!
v0 = 0 ---> initial velocity is zero. Okay.
x0 = 1,28 ---> starting position, to the right of origin. Okay.
 
  • #5
10
0
Oh my!! :redface:

THANK YOU SO MUCH!!! You don't know how confused I was over this 'easy' task... God bless you!
 

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