# Relating acceleration to distance and time

• golya
In summary, the textbook procedure yields a result that is double that of your attempt because vˉ= vf/2 where vf = ∆v assuming constant acceleration.
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?

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.

berkeman and PeroK
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:$$\Delta x = \frac 1 2 a t^2$$$$a = \frac{2\Delta x}{t^2}$$

MatinSAR and golya
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!

berkeman

## 1. How is acceleration related to distance and time?

Acceleration is the rate of change of velocity with respect to time. When an object accelerates uniformly, the distance it covers is related to both the acceleration and the time through the equation $$d = ut + \frac{1}{2}at^2$$, where $$d$$ is distance, $$u$$ is the initial velocity, $$a$$ is acceleration, and $$t$$ is time.

## 2. What is the formula to calculate distance when acceleration and time are known?

The formula to calculate distance when acceleration and time are known, assuming initial velocity is zero, is $$d = \frac{1}{2}at^2$$. If the initial velocity is not zero, the formula becomes $$d = ut + \frac{1}{2}at^2$$.

## 3. How can you find acceleration if distance and time are given?

To find acceleration when distance and time are given, you can rearrange the equation $$d = ut + \frac{1}{2}at^2$$. If the initial velocity $$u$$ is zero, the equation simplifies to $$a = \frac{2d}{t^2}$$. If $$u$$ is not zero, you need additional information about the initial velocity.

## 4. What role does initial velocity play in the relationship between acceleration, distance, and time?

Initial velocity plays a crucial role in the relationship between acceleration, distance, and time. If an object starts with an initial velocity $$u$$, the distance covered is influenced by both this initial velocity and the acceleration. The equation $$d = ut + \frac{1}{2}at^2$$ accounts for the initial velocity, showing that distance is a function of both the initial motion and the acceleration over time.

## 5. Can you derive the equation $$d = ut + \frac{1}{2}at^2$$ from basic principles?

Yes, the equation $$d = ut + \frac{1}{2}at^2$$ can be derived from basic kinematic principles. Starting with the definition of acceleration $$a = \frac{dv}{dt}$$, integrate to find velocity $$v = u + at$$. Then, integrate the velocity equation with respect to time to find distance: $$d = \int (u + at) dt = ut + \frac{1}{2}at^2$$.

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