• Kinematics: In One Dimension and Under Constant Acceleration
  • Sign Convention
  • Velocity Formula
    • Solving for the velocity
    • Solving for the time
    • Solving for the acceleration
    • Solving for the initial velocity
• Next Page

5. Kinematics: In One Dimension And With Constant Acceleration

Kinematics is the description of a particle's motion. In other words, kinematics serves to tell us how a particle moves in time.

Among other things, it will allow us to predict what the velocity of a particle will be at a certain time. It will also allow us to predict the position of a particle at that time.

Once again, for simplicity, we will consider only motion in one dimension. In essence, this is motion along a line. In addition, we will also only consider the case of motion with a constant acceleration. If you are not comfortable with the notion of a constant acceleration, please review this section above.

In the constant acceleration section above, we were able to predict the velocity of a particle without resorting to any mathematical formulae. However, we were unable to predict the position of the particle. In fact, predicting the position of a particle undergoing constant acceleration requires knowledge of calculus. However, do not be alarmed by that fact. All we need are the results obtained from calculus, and we won't actually do any calculus.

Let me quote the formulae that we can derive from calculus concerning constant acceleration and motion in one dimension.

Before we proceed, let me stress one point. If you take a physics course in the future, you will be introduced to many formulae. It is not enough to memorize the formulae, but you must know when they apply and under what conditions you can use them.

For instance, these formulae only apply to the case of a particle moving under constant acceleration. If this condition does not apply to the situation under consideration, then you cannot use these formulae.

Finally, let me say something about the units involved with these formulae. In fact, we have spent an entire section above on units. Since we are now using S.I. units, the following are the appropriate units to use for the quantities in these formulae.

Units
• The unit for distances is meters.
• The unit for velocity is meters/sec.
• The unit for acceleration is (meters/sec)/sec.
• The unit for time is seconds.

If you stay consistent with the use of these units, the answer that you get will be in the appropriate units. For instance, let's say you are solving for the velocity of a particle. If you use the appropriate units for everything else, the answer you get for velocity will be in "meters/sec". If on the other hand, you are solving for acceleration, the answer you get for acceleration will be in "(meters/sec)/sec" if you use the appropriate units for everything else.

1. Sign Convention

   Before we start applying these formulae, let me introduce a sign convention. Since we are working in one dimension, there are only two directions we need to worry about. For instance, if we consider motion in a horizontal direction, the only two directions are left and right. Likewise, if we consider motion in a vertical direction, the only two directions are up and down.

   Mathematically, we can denote the two directions with a sign. The convention that we will use is as follows.

   • Horizontal Motion
     1. Right is (+).
     2. Left is (-).

   • Vertical Motion
     1. Up is (+).
     2. Down is (-).

   This is just a convention, something that we choose to agree upon. We could have designated "up" as (-) and "down" as (+). As long as everyone knows the convention, there is no confusion. In our discussion, we will stick with the convention given above.

   Because we will consider horizontal motion independent of vertical motion, these sign conventions should not cause confusion as long as we make it clear whether the motion is vertical or horizontal.

   Let me give an example of how to use this. For instance, if we say that a car is moving at 30 m/sec to the left, we can just write that the car's velocity is -30 m/sec. Notice how this is considered a velocity now because the sign (whether it is (+) or (-)) tells us the direction of the car's motion. Remember that a velocity includes both speed and direction. The number designates the speed, and the sign designates the direction. Because we have agreed that left is (-) and right is (+), there is no ambiguity here.

   Let me give another example. If a rocket is moving up at the speed of 100 m/sec, we can just write the rocket's velocity as +100 m/sec. If the rocket had been moving downward, then the sign in front of the 100 m/sec would have been negative, (-).

2. Velocity Formula

   

   

   Let me first introduce this formula before giving some examples on how it can be used. As I stated above, it is not good enough to just plug and chug once you know a formula. You must first determine if the formula can be used in your given situation or problem.

   First, let me tell you something which will be true for all the formulae you will encounter here. I will use the above formula to illustrate the point. This formula has four unknown quantities in it, namely, velocity, initial velocity, acceleration, and time. If you are given actual numbers for all of these quantities except one, then you will theoretically be able to solve for the last unknown quantity. For example, if you are given values for velocity, initial velocity, and acceleration, then you should be able to find time with some algebraic manipulation. This works in any combination. If you are given values for time, velocity, and initial velocity, then you should be able to find the acceleration. I hope this is clear.

   The main condition to keep in mind when using this formula is that the object under consideration must be experiencing a constant acceleration. If this is not true, then move along. This formula will not help you. Even if you know all the quantities except for one, this formula cannot help you because the answer it gives you will not be the correct one, the correct answer being the one that reflects what happens in the real world. Remember, we are not just dealing with mathematics here. If the things we describe do not mesh with what goes on in the real world, then we are not doing physics. The whole idea of physics is to predict and to describe the world around us. Contrary to what a lot of beginning students think, physics is not about plugging numbers into formulae. We use mathematics only as a tool to help us describe the world around us. It's not the other way around.

   For example, if someone gave you the velocity, initial velocity, and time and asked you to find the acceleration, this formula would not be able to give you the correct answer if the object was not undergoing a constant acceleration. Certainly, the formula will give you an answer for the acceleration, but it won't give you the correct answer because you have not satisfied the condition under which this formula is valid in the real world. I really want to stress the importance of getting a correct answer versus just any old answer because any old answer just won't do.

   Okay, enough with all the disclaimers and conditions. I was almost beginning to sound like a lawyer.

   The main thing we want to use this formula for is to predict the velocity of an object undergoing a constant acceleration after a certain amount of time. However, as we stated earlier, we can use this formula to solve for any one quantity if values are given for all the others.

   • Example 1
     What is the velocity of an object, initially at rest, if it experiences a constant acceleration of 10 (m/sec)/sec to the right after a period of 3 seconds?

     First, let's try to solve this problem without using the formula above. If you will recall, a constant acceleration means the object will pick up a certain amount of speed over a specified amount of time. The constant acceleration in this case is 10 (m/sec)/sec to the right. Note how we have to specify direction as well as the number because acceleration involves both a number and a direction. What this means, if you will recall, is that the object will pick up an additional speed of 10 m/sec to the right for every second that it undergoes this constant acceleration of 10 (m/sec)/sec. You can read "10 (m/sec)/sec" as "10 m/sec per second". This should better make it clear that the object picks up an additional speed of 10 m/sec for every second it undergoes the constant acceleration. Knowing this, we can find the answer to our problem.

     1. The object is initially at rest.
     2. After 1 second, the object picks up a velocity of 10 m/sec to the right.
     3. After another second, the object picks up another 10 m/sec of speed to the right. Therefore, its velocity is now 20 m/sec to the right.
     4. After another second, the object picks up another 10 m/sec of speed to the right. Therefore, its velocity is now 30 m/sec to the right.

     Therefore, after 3 seconds have passed, the velocity of the object is 30 m/sec to the right. If you count up the time, it should show that 3 seconds have passed. The answer to this example is that the object will move at a velocity of 30 m/sec to the right after undergoing a constant acceleration of 10 (m/sec)/sec to the right for 3 seconds.

     Now that we have solved it the old way, let's try using the formula to see if the formula works.

     
     
     First, we have to make sure that the condition of constant acceleration is satisfied. Because the example explicitly stated that the object was undergoing a constant acceleration of 10 (m/sec)/sec to the right, we can use this formula.

     Second, we must check that all the quantities except for the one we are solving for is given. Since we are solving for v (the velocity of the object after it undergoes the constant acceleration), we need to know the values for the initial velocity, acceleration, and the time that the object undergoes the constant acceleration. If we do not know these three quantities, then we have no hope of solving for v. Since we do know the initial velocity, acceleration, and time under which the object accelerates, we can solve for v using this formula.

     Quantities
     

     1. The initial velocity of the object is 0 m/sec because we stated that it was initially at rest.
     2. The constant acceleration is 10 (m/sec)/sec to the right. This was given as part of the problem.
     3. The time that the object accelerates is 3 seconds.

     Third, we have to make sure that we have the correct signs for everything. Since this problem has to do with the horizontal direction, our sign convention is as follows. Right is (+), and left is (-).

     Quantities with the correct signs
     

     1. Since the initial velocity of the object is 0 m/sec, we do not have to worry about the sign here because it doesn't matter if the 0 is a +0 or a -0. It's all the same.
     2. The constant acceleration must be written as +10 (m/sec)/sec because the acceleration is to the right.
     3. The time that the object accelerates is 3 seconds.

     Finally, all we have to do is plug everything into the formula below to solve for the velocity of the object after it accelerates for 3 seconds.

     

     v = [0 m/sec] + [+10 (m/sec)/sec]*[3 seconds]

     v = +30 m/sec

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     In the second line above, I plugged in the values for each corresponding quantity into the formula. The third line tells us that the answer is +30 m/sec.

     It is very important to notice the sign in the answer because it tells us the direction of the velocity. Because it is (+), it tells us the object is moving to the right.

     Therefore, the answer we get from using the formula is that, after accelerating at 10 (m/sec)/sec to the right for 3 seconds, the object's velocity is 30 m/sec to the right.

     If you compare this with the answer we got above without using the formula, you will see that they are identical. Therefore, the formula does indeed work.

   At this point, you might not see why using the formula is any more convenient. However, there will be cases when using this formula greatly simplifies things. In addition, if I had not chosen such nice numbers, the math required without using the formula would have been much worse. Let me give a couple of more difficult examples to help illustrate the point and to also gain some practice with using this formula.

   Before giving more examples, however, let me just summarize the procedure we used for this formula.

   Procedure for the Velocity Formula
   

   1. Determine if the conditions that apply to this formula are fulfilled, namely that the object is undergoing a constant acceleration.
   2. Make sure that we have values for all the quantities except for the one we are solving for.
   3. Make sure that the signs are correct for any quantities that have a direction, such as velocity and acceleration.
   4. Plug in the values according to the formula.
   5. Finally, solve for the quantity under consideration. This step usually involves a little algebra.

   Now that we know the procedure, we can really just become robots. As long as we follow the procedure, we should be able to get the correct answer. You might be wondering why I didn't just introduce this formula earlier in the constant acceleration section. It certainly would have made our lives much easier, however, I don't think we would have gotten as good of an understanding of what constant acceleration means if we had started with the formula. Ok, enough of that. Onward to more difficult examples.

   • Example 2
     Consider a ball thrown upward with an initial velocity of 20 m/sec. What will it's velocity be after 3 seconds if it undergoes a constant acceleration of 10 (m/sec)/sec downward? This example will hold special meaning later. For now, just enjoy it.

     

     As before, let's try solving it first without using the velocity formula. You should already get an intuitive feel of what this ball should do because we have already discussed what happens when the velocity is opposite to the direction of the acceleration. At first, the ball should continue moving upward while slowing down. After it reaches a speed of 0 m/sec, the ball will then start to speed up while moving downward. If this isn't familiar to you, please refer to this section. Remember, it is often easier to think of the situation using the concept of a force. The answer should still be the same because net force and acceleration are always in the same direction.

     Let's continue. Because the acceleration is 10 (m/sec)/sec downward, the ball should pick up a speed of 10 m/sec downward for every second it undergoes the constant acceleration. In addition, recall that when the velocity and the acceleration are in opposite directions, we have to subtract the 10 m/sec for every second. When they are in the same direction, we have to add the 10 m/sec for every second.

     1. Initially, the ball's velocity is 20 m/sec upward.
     2. After one second, the ball's velocity is 10 m/sec upward. It is still moving up but is slowing down because the acceleration and velocity are in opposite directions.
        (Note that we subtract the 10 m/sec because the velocity and acceleration are in opposite directions.)
     3. Two seconds after the initial time, the ball's velocity is 0 m/sec. From 1 to 2 seconds, the ball is slowing down while moving upward because the velocity and acceleration are in opposite directions.
        (Once again, we subtract the 10 m/sec because the velocity and acceleration are in opposite directions.)
     4. Three seconds after the initial time, the ball's velocity is 10 m/sec downward. After the ball's velocity becomes zero, the ball begins to move downward while picking up speed.
        (Note that we begin to add the 10 m/sec for every second after the velocity reaches zero. Once the velocity reaches zero, we can treat the situation as if the ball were accelerating from rest.)

     Therefore, the answer is that after 3 seconds of undergoing a constant acceleration of 10 (m/sec)/sec downward, a ball thrown upward with an initial velocity of 20 m/sec upward reaches a velocity of 10 m/sec downward.

     Next, let's try using our velocity formula to see if it produces the same result.

     
     

     All we have to do now is to follow the procedure outlined above for the velocity formula. Let's just go through it step by step to avoid any errors.

     1. The condition of constant acceleration is satisfied because it was explicitly stated in the problem.
     2. We are solving for v, the velocity after the ball experiences the constant acceleration. Since we are given values for all the other quantities in the velocity formula, we can use this formula to solve for v.
        
        • The initial velocity is 20 m/sec upward.
        • The acceleration is 10 (m/sec)/sec downward.
        • The time under which the ball experiences a constant acceleration is 3 seconds.
     3. Next, we need to assign the correct signs to the quantities that have directions, namely the initial velocity and the acceleration, using the sign convention above. Since we are dealing with vertical motion, up is (+) and down is (-).
        • The initial velocity is +20 m/sec.
        • The acceleration is -10 (m/sec)/sec.
        • The time is 3 seconds.
     4. Finally, let's plug the values into the velocity formula and get our result.
        
        • v = [+20 m/sec] + [-10 (m/sec)/sec]*[3 seconds]
        • v = -10 m/sec

     The answer is, therefore, that the ball's velocity is 10 m/sec downward. The (-) sign is very important here because it tells us the direction of the velocity is downward. This answer should match the answer we got above without using this formula.

   That's good news. This should give us more confidence in the velocity formula. Let me just give you one more example to work out so you can get better acquainted with using this formula. It won't be different from what you just did, but it will illustrate how useful this formula is.
   • Example 3
     Assume there is a car moving to the left at an initial speed of 14 m/sec. In addition, assume the car is experiencing a constant acceleration of 5.3 (m/sec)/sec to the right. What will its velocity be at 1.7 seconds and at 3 seconds? Keep in mind that both of these times are measured from the initial time. It is 3 seconds after the initial starting time, not 3 seconds after 1.7 seconds have expired. Likewise, it is 1.7 seconds after the initial starting time when the car has a initial speed of 14 m/sec.

     You already have all the skills you need to solve this problem. Don't panic. Just follow the procedure above, and you should get the answer.

     Before you start, however, try to get an idea of what the car will do. You should already know this also. The car's velocity and acceleration are in opposite directions. Knowing this, what should the car do?

     If you stated that the car should move to left while slowing down to a velocity of 0 m/sec and then it should start moving to the right while speeding up, you are correct. Now, try to solve the problem. I will write down the answers for you to compare.

     Answer

     1. After 1.7 seconds, the car's velocity is -4.99 m/sec which means the car is still moving to the left because of the (-) sign. Notice that, as expected, the car is slowing down.
     2. After 3 seconds, the car's velocity is +1.9 m/sec which means it is moving to the right because of the (+) sign. At this point, the car has already slowed down to zero and is just starting to pick up speed while moving to the right.

     If these are the answers you came up with, then you are already pretty comfortable with how to use the velocity formula. If your answers are different, try reviewing the second example above. That should help shed some light on this problem.

   As I stated earlier, finding the final velocity of the particle after it has accelerated for some time is something the velocity formula can be used for. However, there are several other ways of using it if we suitably rearrange the formula. Remember that if we know the values for all the other quantities except for the one we are interested in, then we should theoretically be able to solve for the one quantity we are interested in.

   The procedure for these other variations are similar to the one used above. The only thing that is different is the arrangement of the formula. I will introduce each variation, followed by a brief description.

   One note of importance is the fact that, since these different variations can be derived from the velocity formula, the requirement of a constant acceleration still applies for all of these variations.

   1. Solving for the time
      The variation of the formula to use for this purpose is as follows:

      
      

      • For example, if someone asked how long it took a car to accelerate from 10 m/sec to 35 m/sec while going to the right at a constant acceleration of 5 (m/sec)/sec to the right, this would be the exact variation that you need.

        You would simply plug in the following:

        

        Notice how we still need to follow the sign conventions and assign a sign to any quantities that have a direction.

        After plugging in these values into the formula above, you should get the answer that t = 5 sec. In other words, it takes 5 seconds for the car to accelerate from 10 m/sec to 35 m/sec. Keep in mind that, once again, this formula is valid only for cases where the acceleration is constant.

      • Let me give you one more example. Imagine a ball that is thrown upward with a velocity of 5 m/sec. If the ball experiences a downward constant acceleration of 10 (m/sec)/sec, how long will it take for its velocity to reach 25 m/sec downward?

        Try to solve this one on your own before reading any further. If you get the correct answer, then you are doing very well.

        The answer to this question is that it takes 3 seconds. If you didn't get this answer, try going back and see if you have the correct signs and values for each of the other quantities except for time because we are solving for time here. Carefully read the question to figure out which is the velocity, the initial velocity, and the acceleration. If after doing this you still don't have the correct answer, the following are the values you should have used in the formula above to get the correct answer.

        

        If you didn't get the correct answer, don't get discouraged. Try going back and rereading the above section on how to use the velocity formula, including the procedure on how to use it. The procedure to follow here is very similar.

      The main things to figure out when doing these problems are what is the final velocity and what is the initial velocity. It is usually obvious what the acceleration is. Once you have that figured out, the next thing to worry about is the sign. However, if you follow the sign convention, you should not run into any problems here.

      Finally, this formula can be derived from the velocity formula by simple algebra. For those of you that are interested in seeing the derivation, please refer to this section. If you know algebra, you should be able to do this yourself. If you don't, just remember that the derivation is not as important as how to properly use this formula. In addition, since this formula is derived from the original velocity formula, it can only be used in situations involving a constant acceleration.

   2. Solving for the acceleration
      The variation of the velocity formula to use for this purpose is as follows.

      
      

      I will just give one example in this section. You shouldn't have any trouble if you have already gotten this far. Please refer to this section if you would like to see how this variation is derived from the velocity formula.

      Once again, let me remind you that the condition of a constant acceleration must be satisfied in order to use this formula because it was derived from the velocity formula.

      An example of a circumstance under which this formula can be used is as follows.

      • Assume there is a car initially moving to the right at 10 m/sec. Furthermore, assume it accelerates for 4 seconds and ends up with a speed of 14 m/sec but moving to the left. Assuming the acceleration was constant, what is the constant acceleration the car undergoes?

        Before trying to tackle this problem. There is one thing you can figure out even before using the formula. That thing is the direction of the acceleration. If the car is initially moving to the right and then it ends up moving to the left, what direction does the acceleration have to be in? You should be able to answer this question easily. If you have trouble, think of which direction the force has to be in order to change the car's direction from right to left. Remember that net force and acceleration are always in the same direction because it is the force that causes the acceleration. Therefore, if you can figure out the direction of the force, then you also know the direction of the acceleration.

        If you found that the net force and the acceleration must point in toward the left, you are correct. The reason why we want to find this out is because we can use it as a check to see if the answer we get using the formula is correct. Because the acceleration is to the left, the answer we get should be a (-) number because of the sign convention we are using. If the answer is not negative, then we have done something wrong.

        Using the above formula, try to figure out the answer first before reading any further.

        The constant acceleration the car experiences should be -6 (m/sec)/sec. Notice how there is a negative sign. The (-) sign is very important because it serves to tell us that the acceleration is to the left which is consistent with what we had concluded earlier.

        The values you should have used are as follows.

        

   3. Solving for the initial velocity
      Most of the time, the initial velocity will be one of the given quantities. But for those circumstances when it is the quantity that needs to be solved for, I have included the variant formula for completeness. Once again, since this formula variant is derived from original velocity formula, the condition of a constant acceleration must be satisfied in order for you to use this formula. If you are interested in seeing the derivation of this variant formula from the original velocity formula, please refer to this section.

      
      

      As I stated earlier, this formula is used when the quantity you want to solve for is the initial velocity. Since this is a rare case, let me just give one example.

      • Assume you have a ball that ends up with an upward velocity of 10 m/sec after it experiences a downward constant acceleration of 10 (m/sec)/sec for 3 seconds. What is the initial velocity of the ball?

        Once again, let's try to get an idea of what the ball is doing before we try to solve this problem. Since the acceleration is downward, the net force must also be downward because it is the net force that is causing the acceleration. However, since the ball is still rising (going upward), what can we conclude about the ball's motion? In other words, is it slowing down or speeding up? Once again, this is nothing new. You have already encountered this situation and should already know what the ball is doing.

        Well, since the ball is moving upward against a downward force, the ball must be slowing down. "That's obvious", you might say. And, I would have to agree with you. But, what this tells us is that the initial speed of the ball must be greater than the final speed of the ball because the ball is slowing down. Notice how I used the word, speed, here because velocity has a direction and we are just comparing speeds. In addition, since the ball is slowing down as it rises, the initial velocity of the ball must be in the upward direction. You might still be wondering why we even bothered to think about this. I could go on and on again about how we are doing physics here and that it is just as important to get a general idea of how the ball is moving instead of just being a monkey and plugging and chugging. However, I won't, even though by the very act of not doing so, I have done it. Oh well. The reason why we are doing this is because it will serve as a check to see if we have used the formula properly. If the answer you get for this problem is that the initial speed is less than the final speed, then you should be alarmed and go back to double check your work. This is because we know intuitively that the initial speed must be greater than the final speed. In addition, we must make sure the answer for the initial velocity that we get has a (+) sign because we know the initial velocity is upward. This is because the ball is slowing down as it rises.

        Now, try to find the answer to this problem. I will once again quote the answer and the input values later, but you should attempt this problem on your own first before looking below.

        The answer is the initial velocity is +40 m/sec. What this means is that the ball was thrown upward with an initial speed of 40 m/sec. This is good because we had already concluded above that the initial speed must be greater than the final speed and that the initial velocity must be upward.

        The following are the values that you needed to put into the formula above. Note that the (-) sign in front of the "a" is very important because it tells us the acceleration is downward.

        

   Well, that's pretty much everything about how to use the velocity formula and its variants. If you were able to go through the above sections with relative ease, then you are doing well. If not, take this time to review it again. Sometimes it takes more than one look at the concepts for it to crystallize. In any event, make sure you are comfortable with using the velocity formula before moving on.

   On a final note, you might have noticed that, in every example that involved tossing a ball vertically, the acceleration I used was 10 (m/sec)/sec downward. You might have even said that I had a serious Jones for that specific constant acceleration. However, there was a reason why I chose to do this, but I will leave that as a surprise for later. It's either that or I have no imagination and kept going back to the magical number of 10 again and again. I'll leave it to you to make the call.