# B How to calculate velocity based on distance and time?

1. Nov 30, 2017

### flashsnack

Hi everyone, my first of what I can only imagine will be many questions here.

I’ve been learning some kinematic equations and have been doing pretty well calculating acceleration. The issue I’m having is this. I found this tutorial http://www.dummies.com/education/sc...-and-distance-from-acceleration-and-velocity/ on calculating time and distance from acceleration and velocity:

Imagine you’re a drag racer. Your acceleration is 26.6 meters per second2, and your final speed is 146.3 meters per second. Now find the total distance traveled.

You know the acceleration and the final speed, and you want to know the total distance required to get to that speed. This problem looks like a puzzler, but if you need the time, you can always solve for it. You know the final speed, vf, and the initial speed, vi (which is zero), and you know the acceleration, a. Because vf – vi = at, you know that:

t = vf - vi / a
t = 146.3m/s - 0m/s / 26.6m/s2
t = 5.5s

Now you have the time. You still need the distance, and you can get it this way:

s = vit + 1/2at2

The second term drops out because vi = 0, so all you have to do is plug in the numbers:

s = 1/2at2
s = 1/2(26.6m/s2)(5.5s)
s = 402m

In other words, the total distance traveled is 402 meters

I managed to follow along without a problem. I changed some numbers around and managed to recreate the process without looking at the tutorial. Then I began to wonder if it would be possible to calculate velocity given just distance and time (this may or may not have come up when talking comic books with a friend lol). I tried doing it using the numbers provided in the example but I always end up with a different number and it’s always exactly half and I can’t seem to understand why.

Let’s use the same numbers

Distance = 402m
Time = 5.5s

If I want to calculate the average velocity I believe (correct me if I’m wrong please) it is:

v = d/t

So:

v = 402/5.5
v = 73m/s (half of what was given in the example)

I can’t seem to understand what I am doing wrong. Any help would be greatly appreciated thank you.

2. Nov 30, 2017

### Staff: Mentor

Half of the final velocity given in the example. But since the acceleration is constant, the average velocity is just the average of the initial and the final velocities; and since the initial velocity is zero, the average velocity is just half the final velocity. Which is the same as what you got.

Your calculation, in fact, is useful as a sanity check, since it shows that two different methods of calculating the average velocity give the same answer, as they should.

3. Dec 1, 2017

### lekh2003

You are not doing anything wrong at all, you are just getting a little confused between average velocity and final velocity. Your thought process is correct, but the average velocity is very different from the maximum velocity. To be accurate, it is exactly half of the max velocity assuming constant acceleration.

Imagine a more simple example. You start of at 0 m/s and end up at 10 m/s in 10 seconds. Your acceleration is 1 m/s^2. Your max velocity which is the peak you will reach in the limits of the problem is 10 m/s. Now let us calculate the average velocity by finding the velocity at 10 points in time and diving by ten:
• At 1 seconds: 1 m/s
• At 2 seconds: 2 m/s
• At 3 seconds: 3 m/s
• ...
• At 10 seconds: 10 m/s
Ultimately, if you find the average of these velocities, you will see that the average velocity is 5 m/s. Exactly half.

I hope you understand what is happening here. When you divide the entire distance by the entire time, you are finding the average velocity through the entire race. The max velocity at the end of the race, 10 m/s, is the instantaneous velocity at that time. But the average velocity is different.

The same thing is occurring in your problem of constant acceleration.

4. Dec 1, 2017

### sophiecentaur

Just one word of warning. All these equations ('SUVAT') only apply when the acceleration is uniform (constant) all the time. Very few engines actually produce this. When the acceleration is not uniform (most situations that you could think of), you need much more information to solve the problem accurately.

5. Dec 1, 2017

### flashsnack

Thank you so much everyone, I get it now! It seems so obvious now lol. I really appreciate the help. I'm so glad I found this forum, I look forward to learning a lot more.

6. Dec 1, 2017

### Staff: Mentor

You're welcome! And welcome to PF!

7. Dec 1, 2017

### sophiecentaur

Another quick point.
If an engine is working at constant maximum Power output, the Power will be the Motive force times the Speed. That means the motive force will drop off proportionally with increasing speed for constant Power supplied.

8. Dec 2, 2017

### lekh2003

If I'm right, that requires getting into jerk, jounce, snap, crackle and pop. The humorously named derivatives of position.

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