I have the problem that I need to know how many steps I traveled when I go from velocity $a$ to $b$ in a fixed amount of time. I always know how long it will take me to go from velocity $a$ to velocity $b$. I don't how many steps I will travel in the time I brake to reach velocity $b$. What I need is to predict how far the braking process will take me. The breaking can be linear and non-linear but I think this doesn't matter given that I always know the point in time where I hit velocity $b$?

Steps in this case is an arbitrary distance unit.

This seems like a simple issue but I'm a total physics and math novice. An explanation in layman terms would be greatly appreciated.

E: This is not a homework kind of question if it may seem that way. I need to solve a real world issue here, while I appreciate guidance/full explanation of how and why something gets me to the answer I would appreciate a full answer. I don't know how to ask this more precisely as I simply lack the knowledge to do so.


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Good question. In this type of problem you already know a few variables, so you can figure out the rest using formulas. Let's see what we know (by the way, I"m assuming you already know what speed b is and how much time it takes to reach it from speed a. If you don't please let me know):

distance (d) = ? This is what you want to figure out.

Initial speed (Vi) = a (you already know this).

Final speed (Vf) = b (you already know this).

time taken to go from speed a to speed b = t (you already know this as well).

So the variables we know are: Vi, Vf, and t. The unknown variable is: d

There are a few formulas you can use to find d, but in both you need to know acceleration (a). To do this we use the formula:

vf = vi + at

Now you simply plug in what you know for vf, vi, t, and algebraically solve for a. Remember to include your units and then use them in your calculations (for example, vf would have the unit "steps/second"). Treat the units as variables (like x in algebra), where you can add/subtract like terms.

After solving the above equation for a, you now know acceleration. Here you can choose between two formulas for solving for d (distance):

vf^2 = vi^2 + 2ad


d = vi * t + 0.5a * t^2

You can use either one of these formulas to solve for d. Just plug in what you know for the variables and then algebraically solve for d. In this case I would use the bottom formula, since d is already isolated by itself.

For the future, here are four kinematics formulas that you can use to solve most basic kinematics problems (three of them are ones I've already typed in this post):

vf = vi + at

v_average = (vf + vi) / 2 ......... v_average is just average speed

vf^2 = vi^2 + 2ad

d = vit + 0.5at^2

For these you can always rearrange the variables to solve for a specific unknown.

You could also go to Google Images and just type in "Kinematics formulas". You should find ones that are similar to what I listed. You can almost always rely on these.

If you have any questions on this, please let me know.

  • $\begingroup$ Thank you for this very detailed answer. It was easy to follow. I do have one question. Do these formulas apply for every kind of unit? It makes a lot of sense in my mind right now but I just want to go sure. If I just make up any distance unit and measure how many units I travel per second I can use the kinematic formulas the same way as I would with lets say meter/s? If yes would the same be true for made up time units as long as I'm consistent? F.e. steps/tick instead of steps/s. $\endgroup$ – PTS Mar 24 '17 at 22:08

I assume you know the time duration in which you move from rest to speed a. With that first find the distance covered during the speed a and since you know the time in which the particle moves from speed a to speed b use it to find the distance covered during the process. You have mentioned that you know the time that goes between every steps, as you have found out the distance between speed a and speed b find the total steps covered in the time period and i hope you will get the answer.

  • $\begingroup$ " you know the time in which the particle moves from speed a to speed b use it to find the distance covered during the process" that is my issue though, how do I do that exactly? $\endgroup$ – PTS Mar 23 '17 at 15:41
  • $\begingroup$ Ok, as you have mentioned that the particle is decelarating while moving from speed a to speed b(Braking), find the distance covered during the period by kinematics relation between distance, initial speed and acceleration. I f you don;t know the formula just google for the three kinematics formulae $\endgroup$ – Little-Einstein Mar 24 '17 at 7:25

If you know the braking time period, its all right. In case you don't know, you need the braking acceleration to find it. The braking acceleration is assumed to be anything you wish, thus will be the braking time period. $$t=\frac {\Delta v } a $$ In a layman's sense, if you decelerate from velocity u to v and are willing to take a longer time doing so than before, you'll have to travel a longer distance. Why? The instantaneous velocities you'll have in all instants now will also be same as those you had earlier. But, the duration of instants now will be longer leading to a longer distance traveled.

Similarly, if you undergo a greater change in velocity in a fixed time provided you have the same initial speed, you'll travel a shorter distance. Why? Since, you'll have a greater force acting on you this time, your velocity in all the instants will be lower than in the first case, so you'll eventually travel a shorter distance.

Also, if you begin the journey with a greater initial velocity but still end up with same change in velocity in the same time period, you'll travel a longer distance. The reason is same as above.


Consider you are moving with a certain velocity u and decelerate to speed v. You travel a distance $\Delta x$ in this process and take a time $\Delta t$.

Then, $$\Delta x = \frac {u+v}2 \times \Delta t$$

If your each step has a length $\alpha$ then the Number of steps you take is $$ N= \frac { \Delta x} \alpha $$ So, $$N = \frac {u+v}{2\alpha} \times \Delta t$$ $$N = \frac {2u+\Delta v}{2\alpha} \times \Delta t$$

The equation explains all the stuffs I said earlier.


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