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One Dimensional Kinematics

  • Thread starter Teleknight
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  • #1
Member advised to use the homework template for posts in the homework sections of PF.
So this question has two parts. The first I got without any trouble:
"Two objects move with initial velocity of -8.00 m/s, final velocity of 16.0 m/s, and constant accelerations. (a) The first object has displacement 20.0 m. Find its acceleration."
I used The formula Vf2=Vi2+2aΔx and got an answer of 8 m/s2.
The second part is where I have some difficulty:
"(b) The second object travels a total distance of 22.0 m. Find its acceleration."
Before someone says to solve the same way as part A, I should point out part B says "distance", not "displacement".
If someone could give some insight on which equations to use and if calculus would be required that would be great. If someone wants to actually solve it that would also be fantastic for me to check my work with.

Thanks!
 

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  • #2
SammyS
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So this question has two parts. The first I got without any trouble:
"Two objects move with initial velocity of -8.00 m/s, final velocity of 16.0 m/s, and constant accelerations. (a) The first object has displacement 20.0 m. Find its acceleration."
I used The formula Vf2=Vi2+2aΔx and got an answer of 8 m/s2.
The second part is where I have some difficulty:
"(b) The second object travels a total distance of 22.0 m. Find its acceleration."
Before someone says to solve the same way as part A, I should point out part B says "distance", not "displacement".
If someone could give some insight on which equations to use and if calculus would be required that would be great. If someone wants to actually solve it that would also be fantastic for me to check my work with.

Thanks!
For (a), check your algebra.
What is (-8)2 ?
 
  • #3
Thanks :D. That was a silly mistake to make. Now that that's sorted, any ideas for part B?
 
  • #4
SammyS
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Thanks :D. That was a silly mistake to make. Now that that's sorted, any ideas for part B?
By the way: Welcome to Physics Forums !

Let's say the second object moves a distance d until coming to rest. That's in the negative direction so displacement is -d .

From that point the object moves a distance 22 - d which is in the positive direction, so that is also the displacement.

You get two equations in two unknowns.
 
  • #5
Thanks for the welcome! I'm sorry, I really don't follow. Should I be using the same equation from part A split into two parts? If so, how do I combine them to reach a final answer of a=?
 
  • #6
SammyS
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Thanks for the welcome! I'm sorry, I really don't follow. Should I be using the same equation from part A split into two parts? If so, how do I combine them to reach a final answer of a=?
Yes. Use the same kinematic equation.

Write the two resulting equations. Then it may become apparent what to do. Else, we will give more help.
 
  • #7
When I do that I get 0=64-2ad and 0=-2ad+44a-256, when I set them equal to each other I can reduce to 44a=320 further reduced to a=7.27 m/s2. Does this seem correct? Also, thanks for all the help.
 
  • #8
SammyS
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When I do that I get 0=64-2ad and 0=-2ad+44a-256, when I set them equal to each other I can reduce to 44a=320 further reduced to a=7.27 m/s2. Does this seem correct? Also, thanks for all the help.
Yes.

As a check on consistency, you can then solve for d. The overall displacement is 22 - d. See if that gives the same acceleration using the method of part (a).

(It does.)
 
  • #9
SammyS
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When I do that I get 0=64-2ad and 0=-2ad+44a-256, when I set them equal to each other I can reduce to 44a=320 further reduced to a=7.27 m/s2. Does this seem correct? Also, thanks for all the help.
By the way, I did come up with an alternate solution to part (b).

It still uses distance, d, as defined above. Suppose it takes time, t, to go from -8 m/s to being at rest, 0 m/s. Then it also takes time t to get back to the starting place at positive 8 m/s . The average speed during either of these segments is 4 m/s.

Now it also takes the same amount of time to go from 8 m/s to 16 m/s, but this is with average velocity of 12 m/s, so the object covers a distance of 3d during this segment.

Therefore, 5d = 22 m and thus d = 4.4 m.

You can then use the same kinematic equation to find the acceleration, using any of these time intervals .
 
  • #10
That's a pretty good explanation. Thanks for the help and having patience.
 

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