Constant acceleration motion problem

In summary: I'm actually relieved that I don't have to take that class. It sounded really tough.In summary, the particle moves along a straight line with constant acceleration from 0 to 4s, then changes direction at 4s and moves towards -2 m with a velocity of -2.4 m/s. At 10s, the particle is observed to have reached its final position -2 m away from a chosen reference frame.
  • #1
Bensky
82
0

Homework Statement


A particle moving along a straight line with constant acceleration starts its motion at t = 0. The particle is observed to change direction at t=4s, and when t=10s it reaches the position of -2 m with a velocity of -2.4 m/s measured from a chosen reference frame. Find the position of the particle when it changes direction of motion.

Answer: 5.2 m


Homework Equations



x = x0 + v0*t + (1/2)at^2
v = v0 + at

and possibly: v^2 = v0^2 + 2a(x-x0)

The Attempt at a Solution



I'm not really sure how to go about doing this problem. I'm getting a little confused on the defining variables part even -- this is what I THINK they should be defined as but I'm not sure:
x = -2 m at t=10
x0 = ?
v = 0 m/s at t=4s, -2.4 m/s at t=10
v0 = ?
a = ?

It seems like I'm missing 3 things. I'm thinking either v0 or x0 is 0, but I'm not sure which of those is 0.

I think I'll only be using the first 2 equations since time is involved, and that's really as far as I got on this problem. :( Any help is appreciated, I have a quiz that includes this problem due tomorrow at 1:25 PM eastern time.
 
Physics news on Phys.org
  • #2
Someone can close this -- I figured it out on my own. Used systems to solve this one per a friend's tip. Solution is below for anyone interested:

0 = v0 + 4a
-2.4 = v0 + 10a
==============
2.4 = -6a
a = 2.4/-6 = -0.4

0 = v0 + 4 * -0.4
v0 = 1.6

-2 = x0 + 1.6 * 10 + .5 * -0.4 * 100
x0 = 2

x = 2 + 1.6 * 4 + .5 * -0.4 * 16
x = 5.2 m
 
  • #3
How did you get the -6a?
 
  • #4
anon_tss said:
How did you get the -6a?

I just subtracted the two equations. That's why the 2.4 is positive on the lefthand side and the v0 disappears.

Also, this thread is OLD and has been solved already. Someone can close it as I said in the previous post
 
  • #5
I got a similar question on a quiz due tomorrow, Chang ftw
 
  • #6
anon_tss said:
I got a similar question on a quiz due tomorrow, Chang ftw

Haha nice, you're a VT student too? I had Chang a couple semesters ago...I thought he stopped teaching?
 
  • #7
Ya, second semester at tech. Chang still teaches, I was pretty lucky to get him haha
 
  • #8
anon_tss said:
Ya, second semester at tech. Chang still teaches, I was pretty lucky to get him haha

Yeah, honestly, all the Physics professors EXCEPT for Chang are horrendously bad at VT. Enjoy 2306 next semester...it's a fun time.
 
  • #9
Ya I heard, thanks for the help. I have to figure out the rest of the stuff haha
 
  • #10
anon_tss said:
Ya I heard, thanks for the help. I have to figure out the rest of the stuff haha

No problem
 

1. What is constant acceleration motion?

Constant acceleration motion is a type of motion in which the velocity of an object changes at a constant rate. This means that the object's acceleration remains constant throughout the entire motion.

2. How is constant acceleration calculated?

Constant acceleration can be calculated using the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between constant acceleration and uniform motion?

Constant acceleration refers to a change in velocity at a constant rate, while uniform motion refers to a constant velocity with no change in speed or direction.

4. Can an object have constant acceleration and changing velocity?

Yes, an object can have constant acceleration and changing velocity if the direction of its acceleration is changing. This is known as non-uniform acceleration.

5. How do you solve a constant acceleration motion problem?

To solve a constant acceleration motion problem, you need to identify the known and unknown variables, choose an appropriate equation to use, and plug in the values to solve for the unknown variable. It is also important to pay attention to the direction of the acceleration and velocity, as well as the units of measurement being used.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
885
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
590
  • Introductory Physics Homework Help
Replies
16
Views
333
  • Introductory Physics Homework Help
Replies
5
Views
974
Replies
5
Views
743
  • Introductory Physics Homework Help
Replies
29
Views
6K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
907
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top