Basic dynamics/acceleration question

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Homework Help Overview

The discussion revolves around a dynamics problem involving a 1000 kg car traveling at 27.0 m/s and the force required to stop it over a distance of 70.0 m. Participants explore the implications of constant acceleration and the application of kinematic equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial approach of treating the motion as constant speed and question the validity of this assumption. There is mention of using SUVAT equations and velocity-time diagrams to analyze the problem correctly.

Discussion Status

Some participants have offered guidance on the appropriate equations to use and the importance of understanding acceleration in this context. There is acknowledgment of the original poster's confusion and a general sense of support in clarifying the concepts involved.

Contextual Notes

Participants note that the original poster may not have fully grasped the kinematic equations or the concept of average velocity, which could affect their understanding of the problem. There is a reference to coursework that may not have covered these topics in depth.

Mike_Stokes
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Homework Statement



A 1000kg car is traveling at 27.0 m/s. Determine the force required to stop it in 70.0 m.

The Attempt at a Solution



I attempted to treat this question as if I knew how many seconds it took to stop, and did this:

70.0m / 27.0m/s = 2.59259s

2.59259s / 27.0m/s = 0.096

1000kg / 0.096 = 10414.29N

F = 10.4kN

But then I realized that this is all predicated on the idea that we are moving at 27.0 m/s throughout the whole 70.0m rather than accelerating. Now I have no idea what to do, and both of my textbooks are proving useless. Please help, and sorry if the question is messed up or hard to decipher, this is my first post.
 
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Mike_Stokes said:

Homework Statement



A 1000kg car is traveling at 27.0 m/s. Determine the force required to stop it in 70.0 m.


The Attempt at a Solution



I attempted to treat this question as if I knew how many seconds it took to stop, and did this:

70.0m / 27.0m/s = 2.59259s

2.59259s / 27.0m/s = 0.096

1000kg / 0.096 = 10414.29N

F = 10.4kN

But then I realized that this is all predicated on the idea that we are moving at 27.0 m/s throughout the whole 70.0m rather than accelerating. Now I have no idea what to do, and both of my textbooks are proving useless. Please help, and sorry if the question is messed up or hard to decipher, this is my first post.
You should first understand and become familiar with the SUVAT equations of motion for constant acceleration which you can find in any text to solve for the acceleration, then correctly apply Newton's Laws.
 
"suvat" equations are sometimes called "kinematic equations"
you can also work stuff out from the velocity-time diagrams.

you should have had some coursework about force and acceleration
- if you have not seen v-t diagrams, then you have probably had some work on average velocity.

Using average velocity to do this will make most people here cringe, but some courses do it that way at first.
 
Simon Bridge said:
"suvat" equations are sometimes called "kinematic equations"
you can also work stuff out from the velocity-time diagrams.

you should have had some coursework about force and acceleration
- if you have not seen v-t diagrams, then you have probably had some work on average velocity.

Using average velocity to do this will make most people here cringe, but some courses do it that way at first.

Ahh okay, yeah we've done kinematic equations, but I was having a hard time remembering them, and we never referred to SUVAT at all as an acronym. Thanks so much for the help guys, I finally got the right formula for the question:

a = (Vf^2 - Vi^2)/2D

F = 5.26kN

Thanks for helping my (now seeming stupid) question!
 

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