What is the value of the automobile acceleration?

AI Thread Summary
The discussion revolves around calculating the acceleration of an automobile that starts from rest and travels a distance of 64 meters between two time markers, 8 seconds and 12 seconds. Participants debate the correct approach to find acceleration, emphasizing the difference between average and instantaneous velocity. The book's answer of 1.6 m/s² is questioned, with some arguing that the correct acceleration should be 2 m/s² based on the calculations. The confusion stems from interpreting the problem's conditions, particularly regarding the starting point and the nature of constant acceleration. Ultimately, the participants seek clarity on the formulas and concepts related to constant acceleration in physics.
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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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