How Fast Is a Race Car After Traveling 200 Meters?

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Homework Help Overview

The discussion revolves around a race car that accelerates uniformly from rest at a rate of 4.90 m/s², with the goal of determining its speed after traveling 200 meters. Participants are exploring the appropriate kinematic equations to use in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial attempts to calculate speed using a formula that may not be complete. There are questions about the kinematic equations being applied and the assumptions made regarding time and distance.

Discussion Status

Some participants are providing guidance by prompting others to show their work and check their notes for relevant kinematic expressions. There is an ongoing exploration of different approaches to arrive at the correct formula for the problem.

Contextual Notes

One participant notes the absence of time in the problem setup, which raises questions about how to proceed with the calculations. There is also mention of a borrowed example sheet that could provide additional context for the discussion.

vaironl
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Homework Statement


A race car starting from rest accelerates uniformly at a rate of 4.90m/s^{2} . What is the car's speed after it has traveled 200m?

Homework Equations



Δv = \frac{Change in position}{change in time}

The Attempt at a Solution


v=\sqrt{980m^2/s^2}

=31.3m/s
 
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vaironl said:

Homework Statement


A race car starting from rest accelerates uniformly at a rate of 4.90m/s^{2} . What is the car's speed after it has traveled 200m?

Homework Equations



Δv = \frac{Change in position}{change in time}

The Attempt at a Solution


v=\sqrt{980m^2/s^2}

=31.3m/s
Show your work so that we can help you see where your error is.

How did you get v=\sqrt{980m^2/s^2}\,?
 
What kinematic formula are you using? It looks like you've missed out a constant.
 
SammyS said:
Show your work so that we can help you see where your error is.

How did you get v=\sqrt{980m^2/s^2}\,?

gneill said:
What kinematic formula are you using? It looks like you've missed out a constant.

Well I'm required to find the speed after 200 meters, but I was never given any time.

So I assume... Which is not a good thing to do, multiplying the acceleration * the distanced travel would give me the speed.

But I really can't remember the actual equation.
 
vaironl said:
Well I'm required to find the speed after 200 meters, but I was never given any time.

So I assume... Which is not a good thing to do, multiplying the acceleration * the distanced travel would give me the speed.

But I really can't remember the actual equation.

If you can't remember you should check your notes or textbook! Your formula is close to being a valid kinematic expression,... but it's missing a constant. Do you have a list of the common kinematic expressions?
 
gneill said:
If you can't remember you should check your notes or textbook! Your formula is close to being a valid kinematic expression,... but it's missing a constant. Do you have a list of the common kinematic expressions?

I have the textbook, and I did have an example sheet but my friend borrowed to check her work. I found this expression which seems to satisfy my problem, v2 = vo2 + 2a(X - Xo)

Therefore V2= 0m/s + 2(4.90m/s^2)(200m)

v2= 9.8m/s^2(200m)
v2= 1960m^2/s^2
v=\sqrt{1960m^2/s^2}
v= 44.3m/s
 
vaironl said:
I have the textbook, and I did have an example sheet but my friend borrowed to check her work. I found this expression which seems to satisfy my problem, v2 = vo2 + 2a(X - Xo)

Therefore V2= 0m/s + 2(4.90m/s^2)(200m)

v2= 9.8m/s^2(200m)
v2= 1960m^2/s^2
v=\sqrt{1960m^2/s^2}
v= 44.3m/s

That looks better :smile:
 

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