Relating acceleration to distance and time

AI Thread Summary
The discussion focuses on understanding the relationship between acceleration, distance, and time, particularly for a beginner seeking clarification. The user initially misapplied the formula for acceleration by using distance divided by time squared, leading to an incorrect calculation. The correct approach involves using average velocity and recognizing that average speed is not the same as change in velocity. The textbook's method, which incorporates average speed and assumes constant acceleration, yields the correct result for acceleration. The user ultimately acknowledges their misunderstanding and the importance of using the proper definitions and formulas.
golya
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Homework Statement
Given distance = 402 metres and time = 5.5 seconds, I need to find acceleration.
Relevant Equations
vˉ= vf/2 = (at)/2
I’m an absolute beginner and I need someone to show me where I’m wrong.

Knowing the formula of acceleration ∆v (change in velocity) / ∆t (change in time) where ∆v = ∆x (distance) / ∆t, a common way of relating acceleration to distance is to say a (acceleration) = (distance/time)/time = distance/time^2.

Given distance = 402 metres and time = 5.5 seconds, I need to find acceleration.

Thus I proceeded by calculating a = 402/5.5^2 = 402/30.25 = 13,28 m/s^2. In the same manner I thought I could calculate velocity = 402/5.5 = 73.09 m/s.

However, my textbook reaches a different answer where I don’t understand the thought process.

My textbook proceeds with the formula

s (displacement) = vˉ (average speed) x t (time)

continuing with the formula

vf (final speed) = a x t

deriving

vˉ= vf/2 = (at)/2

Plugging into the above formula s= vˉt, we reach

s = [(at)/2]t = at^2/2

Only now it proceeds to deriving acceleration from displacement and time:

a = 2s/t^2 = 2x402 m / 5.5s^2 = 27 m/s^2

In short, my attempt was using ∆v while their procedure is using vˉ reaching exactly twice my answer because vˉ= vf/2 where vf = ∆v assuming constant acceleration.

But why do they use average velocity instead of change in velocity if a = distance/time^2?

What am I missing?
 
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golya said:
Homework Statement: Given distance = 402 metres and time = 5.5 seconds, I need to find acceleration.
Relevant Equations: vˉ= vf/2 = (at)/2

I’m an absolute beginner and I need someone to show me where I’m wrong.

Knowing the formula of acceleration ∆v (change in velocity) / ∆t (change in time) where ∆v = ∆x (distance) / ∆t, a common way of relating acceleration to distance is to say a (acceleration) = (distance/time)/time = distance/time^2.
This is not right. Average velocity is displacement/time. Velocity is the derivative of displacement with respect to time:$$v_{avg} = \frac{\Delta x}{\Delta t}$$$$v = \frac{dx}{dt}$$
 
PeroK said:
This is not right. Average velocity is displacement/time. Velocity is the derivative of displacement with respect to time:$$v_{avg} = \frac{\Delta x}{\Delta t}$$$$v = \frac{dx}{dt}$$
He is right because the acceleration is constant and the initial speed is 0
 
I think I understand what happened.

My first mistake: distance/time^2 is NOT actually a formula for acceleration but merely an illustration of why acceleration is measured in terms of m/s^2. Therefore the phrase refers merely to units of measurement and not to a formula.

My second mistake: the formula ∆x/∆t (distance/time) does not yield ∆v (change in speed) but vˉ (average speed). This second mistake was the result of the first one.

Therefore the textbook procedure makes sense.
 
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AlexJicu08 said:
He is right because the acceleration is constant and the initial speed is 0
You're wrong as well. Assuming constant acceleration from rest:$$\Delta x = \frac 1 2 a t^2$$$$a = \frac{2\Delta x}{t^2}$$
 
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PeroK said:
You're wrong as well. Assuming constant acceleration from rest:$$\Delta x = \frac 1 2 a t^2$$$$a = \frac{2\Delta x}{t^2}$$
Thank you!
 
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