If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Deriving displacement as a function of time, acceleration, and initial velocity

AP.PHYS:
CHA‑4.A (EU)
,
CHA‑4.A.1 (EK)
,
CHA‑4.A.1.1 (LO)
,
CHA‑4.A.2 (EK)
,
CHA‑4.A.2.1 (LO)
,
CHA‑4.A.2.3 (LO)
Displacement in physics is a vector quantity that measures the change in position of an object over a given time period. Learn how to calculate an object’s displacement as a function of time, constant acceleration and initial velocity. Created by Sal Khan.

Want to join the conversation?

  • orange juice squid orange style avatar for user Syed Haider Hussain Naqvi
    So far what I have studied is that g is positive downwards and negative upwards... you are confusing me by saying that it is negative downwards... please clear this concept!
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Carolina U.
    Why did Sal use average velocity as opposed to just velocity in the equation: displacement = average velocity multiplied by change in time in ?
    (26 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user Kartikeye
    What does force mentioned in the video exactly mean?
    (13 votes)
    Default Khan Academy avatar avatar for user
    • leafers ultimate style avatar for user Ingo
      In physics, a force is any influence that causes an object to undergo a certain change, either concerning its movement, direction, or geometrical construction. It is measured in the SI unit of newtons and represented by the symbol F. In other words, a force can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate, or a flexible object to deform, or both. Force can also be described by intuitive concepts such as a push or a pull. A force has both magnitude and direction, making it a vector quantity.
      (26 votes)
  • leaf green style avatar for user shiv kumar
    when Sal tells at that up is positive and down is negative should it be he other way round,because the work done is against gravity so up will be negative and the 'g' will be positive since its in the same direction?
    (15 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user Robert
      That is a good question. In physics and engineering for example, it doesn't actually matter which direction you select for positive and negative. They are conventions you can set yourself and become "stated assumptions". As long as you are consistant in your direction conventions, it is entirely acceptable to select either direction to be positive. I hope that helps.
      (17 votes)
  • piceratops ultimate style avatar for user Amy
    Does the mass of the Earth ever change?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • female robot ada style avatar for user the.shinyiest
    Sorry that this is such a base question but at why did Sal divide the a times delta t term by 2?
    (8 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Solo
      The delta t at the end was the same delta t at the end of the equation for finding Displacement. Notice how he put all that stuff in brackets? Thats because all the terms in brackets is how we are finding average velocity.
      (6 votes)
  • piceratops ultimate style avatar for user Firedrake969
    If there were no air resistance on Earth, would there be a terminal velocity?
    (5 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Ashiq Ibrahim
      well,(I am aware this is 7 years ago, but incase someone needs it, here you go) I'd say, it depends because an object would not travel at terminal velocity at AIR as there is no air resistance to balance out the weight of the object, so it would accelerate due to gravity. As for an object falling through WATER, or some fluid.. It would travel at terminal velocity as the weight of the object equals upthrust(Assuming Viscous drag negligible) so the answer is yes, there would be terminal velocity, but not for objects falling through air, Basically the medium at which it falls would be whats important in this case.
      Hope this helps someone!
      (2 votes)
  • starky tree style avatar for user amol sharma
    i have a question which i think is related to this topic:
    Two balls of masses m1 and m2 are thrown vertically upward with the same speed u. If air resistance is neglected, they will pass through their point of projection in the downward direction with a speed x. find x.
    please help me understand this question.
    (4 votes)
    Default Khan Academy avatar avatar for user
    • old spice man green style avatar for user Matt B
      Energy lost is equal to the energy gained. It starts with kinetic energy because it has speed u, then all of this energy is transferred to potential energy because it is higher up and is not moving, and then it comes back down and loses all of its potential energy but regains it in kinetic energy. Since there is no air resistance, no energy is lost, therefore the initial speed is also the final speed. Answer: speed _u_
      (6 votes)
  • purple pi purple style avatar for user Alex Hickens
    Why is r the radius of earth?? Shouldn't it be the height in which we throw the ball or rock? Or is Sal talking about something else and I'm just lost...
    (2 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user Sreekar
    We use gravity in place of acceleration. Why is gravity an acceleration rather than a force?
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

What I want to do with this video is think about what happens to some type of projectile, maybe a ball or rock, if I were to throw it straight up into the air. To do that I want to plot distance relative to time. There are a few things I am going to tell you about my throwing the rock into the air. The rock will have an initial velocity (Vi) of 19.6 meters per second (19.6m/s) I picked this initial velocity because it will make the math a little bit easier. We also know the acceleration near the surface of the earth. We know the force of gravity near the surface of the earth is the mass of the object times the acceleration. (let me write this down) The force of gravity is going to be the mass of the object times little g. little g is gravity near the surface of the earth g is 9.8 meters per second squared (9.8m/s^2) Now if you want the acceleration on earth you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. This is accelerating you towards the center of the earth. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying "Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change." And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is all of these terms combined. If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall gravity would be accelerating it downwards, or the force of gravity is downwards. So, little g over here, if you want to give it its direction, is negative. Little g is -9.8m/s2. So, we have the acceleration due to gravity. The acceleration due to gravity (ag) is negative 9.8 meters per second squared (9.8m/s^2). Now I want to plot distance relative to time. Let's think about how we can set up a formula, derive a formula that, if we input time as a variable, we can get distance. We can assume these values right over here. Well actually I want to plot displacement over time because that will be more interesting. We know that displacement is the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)). Right now we have something in terms of time, distance, and average velocity but not in terms of initial velocity and acceleration. We know that average velocity is the same thing as initial velocity (vi) plus final velocity (vf) over 2. (Vavg=(vi+vf)/2) If we assume constant acceleration. We can only calculate Vavg this way assuming constant acceleration. Once again when were are dealing with objects not too far from the center of the earth we can make that assumption. Assuming that we have a constant acceleration Once again we don't have what our final velocity is. So, we need to think about this a little more. We can express our final velocity in terms of our initial velocity and time. Just dealing with this part, the average velocity. So we can rewrite this expression as the initial velocity plus something over 2. and what is final velocity? Well the final velocity is going to be your initial velocity plus your acceleration times change in time. If you are starting at 10m/s and you are accelerated at 1m/s^2 then after 1 second you will be going 1 second faster than that. (11m/s) So this right here is your final velocity. Let me make sure that these are all vector quantities...(draws vector arrows) All of these are vector quantities. Hopefully it is ingrained in you that these are all vector quantities, direction matters. And let's see how we can simplify this Well these two terms (remember we are just dealing with the average velocity here) These two terms if you combine them become 2 times initial velocity (2vi). two times my initial velocity and then divided by this 2 plus all of this business divided by this 2. which is my acceleration times my change in time divided by 2. All of this was another way to write average velocity. the whole reason why I did this is because we don't have final velocity but we have acceleration and we are going to use change in time as our independent variable. We still have to multiply this by this green change in time here. multiply all of this times the green change in time. All of this is what displacement is going to be. This is displacement, and lets see... we can multiply the change in time times all this actually these 2s cancel out and we get (continued over here) We get: displacement is equal to initial velocity times change in time Some physics classes or textbooks put time there but it is really change in time. change in time is a little more accurate plus 1/2 (which is the same as dividing by 2) plus one half times the acceleration times the acceleration times (we have a delta t times delta t) change in time times change in time the triangle is delta and it just means "change in" so change in time times change in time is just change in times squared. In some classes you will see this written as d is equal to vi times t plus 1/2 a t squared this is the same exact thing they are just using d for displacement and t in place of delta t. The one thing I want you to realize with this video is that this is a very straight forward thing to derive. Maybe if you were under time pressure you would want to be able to whip this out, but the important thing, so you remember how to do this when you are 30 or 40 or 50 or when you are an engineer and you are trying to send a rocket into space and you don't have a physics book to look it up, is that it comes from the simple displacement is equal to average velocity times change in time and we assume constant acceleration, and you can just derive the rest of this. I am going to leave you there in this video. Let me erase this part right over here. We are going to leave it right over here. In the next video we are going to use this formula we just derived. We are going to use this to actually plot the displacement vs time because that is interesting and we are going to be thinking about what happens to the velocity and the acceleration as we move further and further in time.