What is the value of the automobile acceleration?

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Homework Help Overview

The problem involves an automobile starting from rest and undergoing constant acceleration while traveling a distance of 64 meters between two marks at different times. Participants are trying to determine the value of the acceleration based on the given information.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the difference between average and instantaneous velocity, questioning how to relate distance traveled to constant acceleration.
  • Some participants express confusion about using the correct formulas for constant acceleration and whether the initial conditions affect the calculations.
  • There is a debate about the correctness of the book's answer and the implications of starting from rest versus the observed motion between the two time marks.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem and questioning the assumptions made about the initial conditions and the nature of the motion. Some guidance has been offered regarding relevant formulas, but no consensus has been reached on the correct approach to find the acceleration.

Contextual Notes

Participants note that the problem states the automobile starts from rest, but there is confusion about the implications of this when analyzing the motion between the specified time intervals. There is also a concern about the accuracy of the provided answer in the textbook.

jtagtp
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An Automobile starting from a rest at t=0 undergoes constant acceleration on a staight line. It is observed to pass two marks separated by 64m, first at t=8 and the second at t=12s. What is the value of the acceleration?


I know how to find the Vel. from this (64m/4s) but i don't understand how I am able to find the acceleration?

The answer in the book is 1.6 m/s^2 but i what to understand how to do it. Thank You

J
 
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1.Your expression for the (instantaneous) velocity is totally wrong; you're mixing up with the average velocity.

2. Ask yourself:
How is "distance traveled in time t" related to constant acceleration?

3. Welcome to PF!
 
Man I am so lost. Ok i relized i used the average velocity, but what i need to do is use the instantaneous velocity? I never did understand inst. vel?

Thanks for welcoming me, i loved physics in high school but so far this college physics is :confused: :mad: :cry:
 
OK, let's take it slow&easy:
What formulas do you know which is about constant acceleration?
 
constant acceleration =(v(t2)-v(t1))/(t2-t1) = chance in v over the change in time.
 
Do you know d= (1/2)a t2 for constant acceleration starting from rest?
 
So i guess i can try and use the formula and switch it around. a=2d/(t^2)
 
ummm but the problems is that this in not from a dead stop.
 
jtagtp said:
An Automobile starting from a rest at t=0 undergoes constant acceleration on a staight line. It is observed to pass two marks separated by 64m, first at t=8 and the second at t=12s. What is the value of the acceleration?


I know how to find the Vel. from this (64m/4s) but i don't understand how I am able to find the acceleration?

The answer in the book is 1.6 m/s^2 but i what to understand how to do it. Thank You

J

Either the answer in the book is wrong, or the question is flawed.

we know that vi = 0

so vf/2 = 64/8 = 16m/s

a = 16m/s/8s = 2m/s^2. If the acceleration is constant then it should be 2m/s^2 the whole time. However, i don't know where the 1.6m/s^2 came from, unless it didn't start from rest...
 
  • #10
Well it starts from a rest but what we are looking is already moving. It start at 0 but what we are looking at is the section from 8t to 12t
 
  • #11
97gtpacecar said:
Well it starts from a rest but what we are looking is already moving. It start at 0 but what we are looking at is the section from 8t to 12t


read it once more...

jtagtp said:
An Automobile starting from a rest at t=0 undergoes constant acceleration on a staight line...

and besides if it's constant acceleration, it should be the same in the second section.
 

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