Relating acceleration to distance and time

[golya]

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?

 

[PeroK]

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:$$v_{avg} = \frac{\Delta x}{\Delta t}$$$$v = \frac{dx}{dt}$$

 

[AlexJicu08]

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

 

[golya]

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.

 

[PeroK]

He is right because the acceleration is constant and the initial speed is 0

You're wrong as well. Assuming constant acceleration from rest:$$\Delta x = \frac 1 2 a t^2$$$$a = \frac{2\Delta x}{t^2}$$

 

[golya]

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!