# Calculating Minimum Possible Acceleration from given v and d

• AndromedaRXJ
In summary, to find the minimum possible acceleration where the average velocity and distance are known, and the acceleration is assumed to be constant, one can use the formula a= 2d/t^2 where d is the distance and t is the time. This is based on the assumption that the path is symmetric and the acceleration is the same for both halves of the journey. The average velocity can then be found by dividing the total distance by the total time, or half the distance by half the time. It is also important to note that the velocity at the midpoint will be twice the average velocity.
AndromedaRXJ
Basically what I want to do is the opposite of finding average velocity from constant acceleration.

I want to know how to find the minimum possible acceleration where the average velocity v is known and the distance d is know. The acceleration is assumed to be constant.

For for example, a ship in outer space departs from position a and arrives at position b in a straight line. We know the departing and arriving time, and we know the distance, so we know the average velocity. The ship accelerated to the half way point and then decelerated until it stopped at it's destination. The acceleration and deceleration has a constant rate. Initial and final velocities at positions a and b are zero.

How do we find the rate of acceleration?

Last edited:
Rate of acceleration in this case is the same as the rate of deceleration:

a= vf - vi / (t/2) = 2vf / t = 2(d/2)/(t/2) / t = 2d / t^2

paisiello2 said:
Rate of acceleration in this case is the same as the rate of deceleration:

a= vf - vi / (t/2) = 2vf / t = 2(d/2)/(t/2) / t = 2d / t^2

But final velocity is unknown. We only have initial velocity, average velocity, distance and time. And we know the acceleration is constant, we just don't know the exact value. It seems intuitively that this is possible to figure out, but are there too many unknowns? I feel that there isn't.

Are you assuming constant positive acceleration to the midpoint, and constant negative acceleration from the midpoint to the end?

If so, I would recommend assuming that the path is symmetric and solving for the positive half path. Then knowing that acceleration is constant, velocity must be linear, so some form of v(t) = at+c.
You also know the average velocity, which allows you to solve for a. Then for the negative path, you will have -a.

AndromedaRXJ said:
The acceleration is assumed to be constant.
Initial and final velocities at positions a and b are zero.
But,
AndromedaRXJ said:
But final velocity is unknown.

Which is true?

But we do know the velocities. If the ship accelerates to the half way point then decelerates to the end then the two must be at the same (but opposite) rate. Therefore the vf = (d/2)/(t/2) = d/t = vavg.

RUber said:
Are you assuming constant positive acceleration to the midpoint, and constant negative acceleration from the midpoint to the end?

If so, I would recommend assuming that the path is symmetric and solving for the positive half path. Then knowing that acceleration is constant, velocity must be linear, so some form of v(t) = at+c.
You also know the average velocity, which allows you to solve for a. Then for the negative path, you will have -a.

Yeah, I'm assuming constant positive and negative acceleration, where both their rates have the same absolute value.

I really just want the absolute value, so maybe I should just discard the second half of the trip where it decelerates. Basically, the ship has initial velocity 0, accelerates at a constant rate, then reaches some given point in space at a known time where we find average velocity. So I want to know the acceleration for that.

For v(t) = at+c, what is c in this? I didn't find this formula on Google.

C is initial velocity. In this case, zero.

RUber said:
C is initial velocity. In this case, zero.

Okay then I presume v is average velocity then, meaning I have everything I need to solve this. Thank you for your help. I'll try doing the problem with given values now and ask more questions relating to this problem if I run into difficulty.

Thank you to everyone else who also replied.

RUber said:
But,Which is true?

Sorry, what I really meant by final velocity was the velocity at the half-way point. Since both halves of the trips were symmetrical. I got confused there.

In my example v(t) is a linear function, not average velocity. However, it is simple enough to find the average of a linear function.

Let's just talk about the first half of the journey. Since the paths are symmetric, average velocity is the same on the first half as it is on the second and overall.
Vavg = average velocity = total distance / total time = half distance / half time.
Let D be distance to midpoint and T be time to midpoint.
Then assuming constant acceleration A, v(t) = At.
The integral of velocity is position, A(T^2)=D.

RUber said:
In my example v(t) is a linear function, not average velocity. However, it is simple enough to find the average of a linear function.

Let's just talk about the first half of the journey. Since the paths are symmetric, average velocity is the same on the first half as it is on the second and overall.
Vavg = average velocity = total distance / total time = half distance / half time.
Let D be distance to midpoint and T be time to midpoint.
Then assuming constant acceleration A, v(t) = At.
The integral of velocity is position, A(T^2)=D.

So is average velocity plugged into v(t) = At, or initial? And with the two equations you posted, is this a system of equations type of problem with the way you expressed it?

Okay so I plugged in actual values where initial v = 0 m/s, t = 60 sec, distance = 600 meters, so average velocity = 10 m/s. So I plugged in the values to that equation.

v(t) = At

10(60) = A(60)

600 = A60

10 = A

So acceleration is 10 m/s^2? Is this right?

That can't be right. If you are accelerating at 10m/s^2, you will be at 10 m/s after 1 sec, 20 m/s after 2 secs, ... 600 m/s after 60 secs. Does that sound like it would average out to 10 m/s for an average velocity?

v(t) means the velocity at time t, not Vavg*t, as you did above.

If velocity is linear and starts from zero, its average is at its midpoint, and at its final point it is twice the average.

Draw a picture, Acceleration is the slope of the velocity when you graph it. What is the slope of a line that has the properties v(0) = 0 and v(60) = 20?
Also, I had a typo above, A*T^2/2=D, not AT^2 = D. Either of these methods will get you to the correct answer for A.

RUber said:
That can't be right. If you are accelerating at 10m/s^2, you will be at 10 m/s after 1 sec, 20 m/s after 2 secs, ... 600 m/s after 60 secs. Does that sound like it would average out to 10 m/s for an average velocity?

v(t) means the velocity at time t, not Vavg*t, as you did above.

If velocity is linear and starts from zero, its average is at its midpoint, and at its final point it is twice the average.

Draw a picture, Acceleration is the slope of the velocity when you graph it. What is the slope of a line that has the properties v(0) = 0 and v(60) = 20?
Also, I had a typo above, A*T^2/2=D, not AT^2 = D. Either of these methods will get you to the correct answer for A.

So all I really need to remember is the final velocity is twice the average velocity for this type of problem, and from there, I know the time it took to reach that velocity in order to figure out the acceleration.

I wasn't too clear on the graphing method, but for A*T^2/2=D

A*60^2/2=600

A*3,600/2=600

A*1,800=600

A = 6/18 = 1/3

A = 0.3333333333 m/s^2

I got the same answer with this.

http://www.smartconversion.com/unit_calculation/Acceleration_calculator.aspx

So I think I got it right now. Thank you very much for your help!

RUber

## 1. How do you calculate the minimum possible acceleration?

The minimum possible acceleration can be calculated by dividing the change in velocity (v) by the change in distance (d). This can be represented by the formula a = Δv/Δd.

## 2. Why is it important to calculate the minimum possible acceleration?

Calculating the minimum possible acceleration allows us to understand how an object's velocity changes over a given distance. This information is crucial in various fields of science, such as physics and engineering, as it helps us to understand and predict the motion of objects.

## 3. Can the minimum possible acceleration be negative?

Yes, the minimum possible acceleration can be negative. This indicates that the object is slowing down or decelerating. A positive minimum possible acceleration indicates that the object is speeding up or accelerating.

## 4. What are the units for minimum possible acceleration?

The units for minimum possible acceleration are typically meters per second squared (m/s²) in the metric system or feet per second squared (ft/s²) in the imperial system.

## 5. How does the distance affect the minimum possible acceleration?

The distance directly affects the minimum possible acceleration. The larger the distance, the smaller the minimum possible acceleration will be, and vice versa. This is because a smaller acceleration is needed to produce the same change in velocity over a larger distance compared to a smaller distance.

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