Physics Homepage

Solving Motion Problems

When it comes to solving motion problems it is important to realize tat there are two ways to this. The first and better way is to reason your way through the problem from start to finish. By doing so you will make sure that your answer makes sense. The second way to solve these problems is by plugging the numbers into an equation and solving that equation for the unknown. In class I will always solve a problem both ways because they both will give you the right answer. But you should always try to solve a problem by reasoning your way through it rather then using equations.

Let’s solve two problems here:

A child slides down a constant slope on a toboggan.  When the child passes her older brother, he estimates her speed at 2 m/s.  When she passes her younger brother 10 seconds later, he estimates her speed at 10 m/s.  Estimate how far apart her brothers are standing.


Let’s see what we get by using the equations. No doubt you will want to plug the numbers into the third equation, we are looking for the distance, we can find the acceleration and we have the time t. After you find the acceleration a=v/t = 8/10=0.8 m/s/s this seems straightforward, plug in and solve. If you were to do this you would get 40 meters. If you had just used the equations this would probably be your answer and you would have no way of knowing whether this is correct. But the equation d=1/2a t2 cannot be used for this problem, it only works when the object, the cart in this example, starts from rest, which it does not here.

Let’s reason: the cart starts at 2 m/s and ends at 10m/s, it speeds up. It does so for 10 seconds. We could find how far it goes if we knew how much distance it covers each seconds, that is, if we knew the speed. We have two speeds. Can we use either of them? No, but we can find the average speed, which is 6 m/s (initial plus final divided by 2 – if this is not clear, talk to me).  That means the cart covers 6 m in each second and since it is going for 10 seconds it will cover 40 m. Since we reasoned our way through the problem we are confident that it is correct.

Second problem:

A baseball is thrown upward at 20 m/s. At what time is the ball...
a)         30 m above the point at which it was released?
b)         20 m above the point at which it was released?
c)         5 m above the point at which it was released?

Forget the equations and just remember what velocity and what acceleration is. Velocity is how much distance is covered in a second and acceleration is how much the speed changes in a second. The acceleration in this example is gravity, which means that the ball changes it speed by 10 m/s every second. Let’s start with that. Since the ball is thrown upward it will slow down just by that amount – 10 m/s every second. After one second the ball’s speed is 10 m/s and after 2 seconds it will be 0. That means it has reached the turnaround point. The most important thing to realize here is that we have to use the average velocity between the seconds. How high does the ball go anyway? The average velocity between start (20 m/s) and finish (0) is 10 m/s (initial plus final divided by 2 – if this is not clear, talk to me). The ball is going at an average velocity of 10 m/s for 2 seconds – it covers 10 m every seconds and does so for 2 seconds, that means that it will go 20 meters high. Without using any equation we have solved b) and c). The answer for b) is after 2 s and the answer for a) is “never, because it does not go that high”.

What about 10 meters? How high is the ball after 1 s? Its speed will be 10 m/s and the average speed will be 15 m/s (20+10 divided by 2). The ball is going at an average speed of 15 m/s for 1 second, therefore it covers 15 meters in the first second – too much. You could now do the same finding the speed after ½ second and you will find it is still too much. But now we are shooting in the dark; it’s “trial and error”. The last part is difficult.

Let’s see what we get by using the equations. Remember the three equations: v=d/t, a=v/t and d=1/2a t2 . Skip through a) and b) and tackle c). No doubt you will want to plug the numbers into the third equation, we have the distance (5m) we have a = 10m/s/s and we are looking for t. It seems straightforward, plug in and solve. If you were to do this you would get 1 second. If you had just used the equations this would probably be your answer, but if you reasoned your way though the problem like we did above, you will realize that this cannot be right. After one second the ball is 15 m high.

You might not have been able to solve the problem by reasoning alone, but simply using the equations will most likely yield the wrong answer. DO you realize why we got a wrong answer? Think about it before reading further.

The equation d=1/2 a t2 cannot be used for this problem since it only works when the object starts from rest. That’s the problem with equations, you need to know what they mean and that is at sometimes very difficult.

However, we can still sue the equation, but we have to rewrite the problem. Since the equation can only be used when the bal starts from rest we need to look at the motion when it starts from rest – at the turnaround point. Rather then answering the question “When does the ball reach 5 m?”, we will answer the question “When will it have dropped 15 m?”. Since the ball goes 20 m in the air, it’s the same thing. Now we can use the equation d=1/2at2, where d=10 m, a =10m/s/s and we are looking for t. Solving it gives us roughly 1.7 s – 1.7 s after the turnaround point the ball will be 5 m above the ground. Since the motion is symmetrical and it takes 2 s for it to go up, the ball will be 5 m after 0.3 s. 

This is a difficult problem and you might not get part c) right at all. But by reasoning and writing down your thoughts you will get partial credit, which could amount to a lot in this case. If you reason that you cannot use the equation but don’t know how to proceed you might even get full credit. You won’t get any credit if you just plug in the numbers into the wrong equation.


Looking at distance versus time graph allowed us to find the operational definition of speed, namely that it is the distance covered in a certain time interval. In the power point I talk about what would happen if the axis were flipped and we were discussing a time versus distance graph. It is after all just a convention to put the independent variable on the x-axis. In this case we could also find the slope, but it would mean something very different; it would be “how ler, but it is useful nevertheless, when splits are given in cross-country for example.



One thing to keep in mind is that a graph does not represent a picture of the motion, which is evident in the example of the picture below. Describe the motion of biker in this graph. I am sure you will have said that he is going uphill fieriest, then staying on top of the hill and then turning around. However, this is a graph not a picture. All we can tall form the graph is the speed of the biker not where he is in France or at the North Pole, or whether he




Website maintained by Volker Krasemann.