- #1

golya

- 4

- 3

- 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?

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?