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Kinematics Part 4: Practice Problems and Strategy


Professor Dave again, let’s practice solving kinematics problems. We’ve looked at the motion of objects in one and two dimensions, and hopefully the examples we discussed made perfect sense. But many students find that they understand the examples when they are explained but have trouble figuring out what to do on their own during homework or on a test. The solution to this is to simply practice a lot and use critical thinking skills. We will look at a handful of additional examples now and as we go through these the key will always be to ask the following questions: Number one: what data do I have and what variables are used to represent this data? Number two: what is it that I am trying to find out? Three: what equation can I use that will allow me to plug in known values for every variable except the one I want to solve for? Notice that each of these three equations is missing one of the common variables. This one is missing position this one is missing final velocity and this one is missing time. This allows us to do calculations even when there is an unknown parameter. And four: once you arrive at your answer does it make sense logically or is it way too big or too small to be reasonable? If we are able to think critically in this manner we will find that many questions in kinematics become relatively trivial to answer. Let’s give one a shot. Say you toss a ball directly upward with an initial velocity of 5 m/s. What will be the maximum height of the ball? Now as we discussed let’s first decide what we know. We know the acceleration due to gravity, we know the initial velocity, here’s the clever part we also know the final velocity because at the peak of the trajectory the velocity will be 0. We are solving for the position associated with the ball when it will be at its highest point and at that point the velocity is precisely zero before it then starts moving in the negative direction. So which equation has all the things we know plus the thing we want to solve for? That would be this equation where final velocity squared equals initial velocity squared plus 2ax. This is convenient because time is not present, since we have no value for time. Let’s plug in the numbers, solve for x and we should get about 1.28 meters as the maximum height of the ball with respect to the point of release. If we wanted to know its height above the ground we would just add whatever distance there was between the point of release and the ground. Now let’s say we want to know it’s hang time or how long it is in the air from the moment you throw it to the moment you catch it at the same point of release. To do this we can use the following equation and solve for the time elapsed to reach the maximum height and then simply double it as falling will take exactly the same amount of time as ascending. We can use this equation because we know both the initial and final velocities and the acceleration plugging in those values and solving for t gives us 0.51 seconds. That’s how long it takes to reach the maximum height. SImply double that and we get 1.02 seconds as the total hang time. Let’s try one more. Say a helicopter is delivering a care package to a remote location. The helicopter is traveling horizontally at 67 meters per second and is at a height of 120 meters when the package is dropped. How long before the package hits the ground and how far will it travel in the horizontal direction? So the first thing we need to recall is that the vertical and horizontal motion are independent of one another. The package will take the same amount of time to hit the ground whether there is horizontal motion or not so we can isolate the Y direction to find the time elapsed in the air. We know that the position will be negative 120 meters since it will travel that far in the negative direction, initial velocity in the y-direction is zero since the helicopter has no vertical motion, and acceleration is known. Let’s plug these into this equation and solve for t which will be about 4.9 five seconds. Now that we know how long the package is in the air it is a trivial matter to calculate how far it will travel in the horizontal direction. We simply use the definition of displacement. It is traveling with the same velocity as the helicopter the entire time it is in the air since there is no acceleration operating in the horizontal direction so 67 meters per second times 4.95 seconds equals about 332 meters. So remember that when solving problems of this nature always ask yourself what information you have. Also remember that sometimes you may have information that is not explicitly listed in the question. If something is moving from rest the initial velocity is zero. At the moment that an object changes direction with respect to its vertical motion the velocity is also 0. Make sure you think critically in this regard. Then ask yourself what the question is looking for and select an equation that will allow you to solve for this unknown value. Let’s check comprehension. Thanks for watching, guys. Subscribe to my channel for more tutorials, support me on patreon so I can keep making content, and as always feel free to email me:

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