Calculating Acceleration for a Car on a Circular Track

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Homework Help Overview

The problem involves calculating the total acceleration of a car moving on a horizontal circular track with a radius of 80 meters. The car accelerates from rest to a speed of 100 km/h over 10 seconds, and the question focuses on determining the magnitude of total acceleration 8 seconds after the start.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the distinction between tangential and normal acceleration, with some expressing confusion about the definitions and calculations involved. There are attempts to break down the problem into parts and questions about how to apply relevant equations.

Discussion Status

Some participants have provided guidance on using the correct time frame for velocity calculations and the need to consider both tangential and centripetal accelerations. There is an ongoing exploration of how to set up the problem, including vector considerations and the application of formulas.

Contextual Notes

Participants note the challenge of understanding the relationship between tangential acceleration and angular motion, as well as the implications of uniform acceleration assumptions. There is also mention of potential discrepancies in calculated values and the need for clarification on vector components.

krnhseya
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Homework Statement



car is on a horizontal circular track of 80m radius and car starts from rest to 100km/h in 10 seconds. what's the magnitude a of ttotal acceleration of the car 8 seconds after start?


Homework Equations



normal acceleration = v^2/r

The Attempt at a Solution



convert v to m/s and r = 80. then i found normal acc.

i just have no clue on how to do this...class notes and book reading don't help me at all.
what are things i need to consider?
thank you
 
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there are two accelerations:
tangential, and normal.

dividing this problem into two parts might help you.
 
tangential is what's messing me up...
books says it's equal to v with dot above it (derivative)
or also equal to r*theta(with dot above as well)

i don't get what they actually mean or how to get it...
 
krnhseya said:
tangential is what's messing me up...
books says it's equal to v with dot above it (derivative)
or also equal to r*theta(with dot above as well)

i don't get what they actually mean or how to get it...

well, I am assuming here that tangential a is uniform.
Just ignore them, I guess you really don't have an equation to differentiate.

just use that simple for: a = vf-vi/t
10 seconds, and assume that it's same for 8 seconds.

I am pretty sure that this would be correct
 
Yes, you have found the acceleration. But the velocity that you use in the a = v^2/r equation must be the velocity of the car at t = 8 s, NOT at t =10 s, or Vf. You must use the acceleration you just found to find the velocity of the car at t = 8 s.

Also, the total acceleration is the vector sum of the tangential and centripetal accelerations.
 
well can i use v = vo+at ? considering that a is constant and vo is zero(?)
how can you use vector to do this problem...i don't know how to set up the vector...
no coords :(

[edit] well i can sum up and get final vectors i think...only problem is how to get normal acc. and tang acc. in vector form...
 
Last edited:
alright, here's what i have...

v = vo + at (given that vo = 0, v = 100km/h, t = 10)

i get 2.78m/s^2 as acceleration.

v = vo+2.78t (given that vo=0, t = 8)

i get v = 22.2m/s^2

an = v^2/r (given that v = 22.2 and r = 80)

i get an = 6.17m/s^2 and it should be NEGATIVE.

for the vector sum up part...since they are perpendicular to each other, i just find magnitude of two acceleration which is...

sqrt(((-6.17)^2)+(2.78^2)) = 6.77m/s^2

answer is right but just confirming if this is how i should do it.

thank you so much!
 
Last edited:
Yes, that looks right to me.
 
  • #10
ok thank you very much :)
 

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