Solving for Acceleration in a Simple Kinematics Problem

  • Thread starter Thread starter princesssuckz
  • Start date Start date
  • Tags Tags
    Acceleration
Click For Summary

Homework Help Overview

The discussion revolves around a kinematics problem involving a vehicle that accelerates from rest to a final velocity of 14 m/s over a distance of 26 m. Participants are attempting to determine the acceleration of the vehicle using appropriate equations of motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various equations related to constant acceleration, with some questioning their understanding of the formulas and the variables involved. There is also confusion regarding the initial and final velocities.

Discussion Status

The conversation is ongoing, with participants providing guidance on which equations to consider and clarifying the meanings of the variables. There is a recognition of misunderstandings, particularly regarding the initial conditions and the application of the equations.

Contextual Notes

Participants express uncertainty about the equations and their application, noting that some concepts have not yet been covered in class. There is a focus on ensuring that the correct values are used in the equations.

princesssuckz
Messages
13
Reaction score
0
acceleration problem (HELP!:(()

Homework Statement


A vehicle accelerates to 14 m/s after being stopped at a red light. The vehicle covers 26 m to reach its final velocity. How fast was it accelerating?


HOW DO I EVEN SOLVE THIS? WHAT EQUATION AM I GONNA USE?! PLEASE HELP.:(
 
Physics news on Phys.org


What equations do you have?
 


vela said:
What equations do you have?

umm.. acceleration is equal to displacement over time squared.
 


I'm afraid you're going to have to do better than that. Look in your textbook for equations for constant acceleration. Those are the ones you want to choose from.
 


vela said:
I'm afraid you're going to have to do better than that. Look in your textbook for equations for constant acceleration. Those are the ones you want to choose from.

well the only equation here related to acceleration is a=v-vo/delta t
 


Welcome to PF, your highness princesssuckz! :smile:

Try:
[tex]v_f^2 = v_i^2 + 2 a \Delta x[/tex]
Can you make sense of this formula?
 


I like Serena said:
Welcome to PF, your highness princesssuckz! :smile:

Try:
[tex]v_f^2 = v_i^2 + 2 a \Delta x[/tex]
Can you make sense of this formula?

no not really. we haven't discussed this in class yet. the teacher sort of just threw it at us.
 


princesssuckz said:
no not really. we haven't discussed this in class yet. the teacher sort of just threw it at us.

Ah, well, what about this formula?
[tex]x=x_o + v_o t + {1 \over 2} a t^2[/tex]
 


I like Serena said:
Ah, well, what about this formula?
[tex]x=x_o + v_o t + {1 \over 2} a t^2[/tex]

i don't get it. why is there an x?
 
  • #10


princesssuckz said:
i don't get it. why is there an x?

Because the problem statement says: "The vehicle covers 26 m".
This is "x".
 
  • #11


oh now i get it. :smile: i feel really stupid.
 
  • #12


princesssuckz said:
oh now i get it. :smile: i feel really stupid.

Uhh? :confused:

So you solved the problem?
 
  • #13


sorry. i tried but no. could you please walk me through it. :(
 
  • #14


So which equations do you have now?
And what do you already know about the numbers in them?
 
  • #15


I like Serena said:
So which equations do you have now?
And what do you already know about the numbers in them?

just the ones you gave. i tried the first one. but i don't think it's right.
v2f= 14m/s2+2(9.8m/s) (26 m)
:confused::cry:
 
  • #16


princesssuckz said:
just the ones you gave. i tried the first one. but i don't think it's right.
v2f= 14m/s2+2(9.8m/s) (26 m)
:confused::cry:

Ah, well, let me explain:

[itex]v_i[/itex] is the initial speed. What is it?
[itex]v_f[/itex] is the final speed. What is it?
[itex]a[/itex] is the acceleration you have to calculate - it is not gravity!
[itex]\Delta x[/itex] is the change in position - you have that right.
 
  • #17


I like Serena said:
Ah, well, let me explain:

[itex]v_i[/itex] is the initial speed. What is it?
[itex]v_f[/itex] is the final speed. What is it?
[itex]a[/itex] is the acceleration you have to calculate - it is not gravity!
[itex]\Delta x[/itex] is the change in position - you have that right.

[itex]v_i[/itex]= 14 m/s
i don't know the [itex]v_f[/itex]. :frown:
 
  • #18


princesssuckz said:
[itex]v_i[/itex]= 14 m/s
i don't know the [itex]v_f[/itex]. :frown:

Initially the vehicle is in front of a red light.
At this time the vehicle has its initial speed.

After accelerating over a distance of 26 m, the vehicle achieves its final speed.

Try again?
 
  • #19


I like Serena said:
Initially the vehicle is in front of a red light.
At this time the vehicle has its initial speed.

After accelerating over a distance of 26 m, the vehicle achieves its final speed.

Try again?

hmm..so the [itex]v_i[/itex]= 0
and the [itex]v_f[/itex]= 14m/s

right?:confused:
 
  • #20


Right! :smile:
 
  • #21


I like Serena said:
Right! :smile:

now what?:frown:
 
  • #22


princesssuckz said:
now what?:frown:

Fill in the numbers in the formula again.
You have all the numbers except "a".

Then solve the equation for "a" and you have your answer.Can you write down the formula again with the symbols replaced by the numbers as far as you know them?
 
  • #23


I like Serena said:
Fill in the numbers in the formula again.
You have all the numbers except "a".

Then solve the equation for "a" and you have your answer.


Can you write down the formula again with the symbols replaced by the numbers as far as you know them?

sorry if I'm being slow here. so should it be like this 14 m/s2= 0+2a(26 m) ?
 
  • #24


Exactly! :smile:

Can you solve "a" from this equation?
 
  • #25


I like Serena said:
Exactly! :smile:

Can you solve "a" from this equation?

[tex]{14 {m \over s^2}-26m \over 2}= {2 a \over 2}[/tex]

[tex]{14 {m \over s^2}-26m \over 2}= a[/tex]

[tex]{-12 \over 2} = a[/tex]

[tex]-6 = a[/tex]

it's wrong, right?:frown:
 
  • #26


Yeah, sorry, it's wrong.

Let's go back to the equation:
(14 m/s)2= 2a(26 m)

From here you cannot subtract 26 m to bring it to the other side.
(You could already see that this wouldn't work, because the units (m/s)2 and m do not match, which they must for subtraction.)
You need to divide by 26 m to bring it to the other side.

Can you do that?
 
Last edited:

Similar threads

Replies
18
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
5
Views
900
  • · Replies 3 ·
Replies
3
Views
15K
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K