In physics (as well as in life), knowing how to best attack a problem will bring the best results. Coming up with the solutions for physics problems has led to the creation of great stuff like the fire you create when lighting a match to create the bonfire you need for making s’mores. What if I told you your television was developed using physics? That’s right. Advances in nuclear physics or electromagnetism directly led to newer products that have caused big changes in society including domestic appliances, computers and your television set.
Here is a course entitled Physics Essentials 1 that shows you all the fundamentals. Basically, knowing how to come up with new solutions using physics tends to make life a little better than it used to be. Constant practice with physics problems will help you get the most out of your physics class and be a great foundation for more advanced subjects later in in life.
From the ancient Greek word “phusike,” Physics literally means ‘knowledge of nature.’ It involves the study of matter and the way it moves through time and space. A natural science, it also involves related concepts of force and energy. In broader terms, it tries to understand how the universe behaves by doing a general analysis of nature, which apparently, even applies to video games. This article entitled Construct 2 Tutorial: Understanding Physics enables video games to have real world properties using principles of physics.
One of the oldest academic disciplines, physics includes astronomy. Along with chemistry, it is also a part of natural philosophy, biology and branches of math. Here is a course called Astronomy for VCE Physics that involve your own observations of the night sky. Another course on astronomy tells you about the most recent astronomical discoveries.
Physics is About Solving Problems
The whole point about physics is, you guessed it, about problems and getting to their solutions. There may be major principles and themes that are emphasized but really, it all boils down to solving problems with accuracy. When faced with a problem in physics, first ask yourself what the problem is really about. Next, identify what it is you need to find. Decide which existing information you can use to solve it, or which principles of physics. You may also want to think back: Was there a time when you encountered a problem similar to this one? Apply the appropriate information for solving the physics problem you are faced with. Once you come up with a number, check to see if it makes sense.
Using Kinematic Equations to Find Solutions
The motion of objects are represented and described by equations called Kinematic equations. With the motion of objects, there are various quantities associated including time, acceleration, speed and velocity as well as distance and displacement. Knowing about each of these will give you information about the motion of objects. For instance, if a motorbike moves with a velocity constant at 22 m/s for twelve seconds north, with a distance of 264 meters, then there is a full description of the bike’s motion. All the information seems to be given. However, what you don’t know is the displacement your bike would go through if you skidded to a stop. It is during these instances that you can use principles of physics and Kinematic Equations to find out this kind of information. In other words, these equations can be used as guide to determine unknowns about the motion of an object.
d= vi + vf / 2 * t
vf = vi + a * t
vf2 = vi2 + 2 * a * d
d = vi * t + ½ * a * t2
As you can see, there are various symbols involved, each one with its own meaning:
a = acceleration
t = time
d = displacement
v = velocity
vi = initial velocity
vf = final velocity
When you come across physics problems such as the ones listed below, you can use any of the four equations that will help you determine information about the motion of an object based on existing information. Now how convenient is that?
Solving Physics Problems
The equations above can help you determine information you don’t know about the motion of an object. This method involves using strategies for problem solving that will be utilized throughout the course. Involved in the strategy are these steps:
- Constructing a diagram of the situation using information
- Identifying and listing given information in the form of variables
- Doing the same for the information that is unknown.
- Identifying and listing which of the equations you plan to use to find out information you don’t know based on the information you do know
- Substituting variables that are known into the equation and using the right steps of algebra to solve for information that is not known.
- Ensuring correct and reasonable math answers by checking your work.
Here are a couple of examples about solving physics problems that how you how you can use equations to predict moving object parameters that are unknown. As long as there are some known parameters, the values remaining can be found out. You can use the same strategy for other types of motions such as free-fall. You will see in several practice problems below how this applies as well.
To apply this strategy, here are a few physics problems that you could practice with:
Mike was in his car waiting at a traffic light. When it turned green, finally, he sped up with a time of 4.10 seconds and at a rate of 6 m/s2. Find out what the displacement of Mike’s car was at this time.
To solve this problem, you need to construct a diagram of information of the actual circumstance. Next, find a strategy that will identify unknown information. Note that the value vi can be inferred to be zero since initially, Mike’s car was at rest. The car’s acceleration (a) is 6 miles per second with a 4.10s time. Next, list all the information that is unknown. This physics problem asks about what the car’s displacement is. The unknown information is thus d:
a =6 m/s2
t= 4. 10s
vi = 0 m/s
Find the right equation that will give you the unknown information. In this case, it is this one:
Vi * t + ½ * a * t2 = d
The next thing to do would be to substitute all the values known into the equation using all the right steps of algebra for solving for the information you seek:
0.5*(6.00 m/s2)*(4.10 s)2 + (0 m/s)*(4.10 s) =d
0.5*(6.00 m/s2)*(16.81 s2) + (0 m) =d
50.43 m + 0 m =d
50.4 m =d
This solution tells you that the car will be traveling at a 50.4 meter distance. The final step is to find out if your answer is both accurate and reasonable. This looks like a reasonable enough value. If you substitute the calculated value back into the equation, you will find that indeed, it evens out.
Sally is coming near a traffic light moving at a +30 m/s velocity. The light turns yellow and she begins skidding to a full stop. If her acceleration becomes -8 m/s2 find out the car’s displacement during the process of skidding. Keep in mind that the acceleration and velocity vectors of direction are noted by a positive or a negative sign.
To solve this problem, you need to begin by construction a diagram of information from the actual circumstance. Next, identify and list the information you do know in the form of variables. Note that the value vf can be inferred to be zero miles per second (0 m/s) since Sally’s car stops completely. The car’s initial velocity or vf is +30 since she is running at normal speed. The car’s acceleration (a) is given at – 8.00 m/s2. Paying careful attention to the signs of negative or positive is important, by the way. Next, list the information you want to find out. In this case, you want to find out the car’s displacement, which is indicated by the unknown quantity: d. Here is the solution:
a = 8 m/s2
vf = 0 m/s
vi = +30.0 m/s
Next, identify the equation you need to use to find d. As you can see, the right equation to use is this one: vi2 + 2 * a * d = vf2
Substitute all the values known into the equation and solve using the correct steps of algebra:
(30.0 m/s)2 + 2*(-8.00m/s2)*d = (0 m/s)2
900m2/s2 + (-16.0m/s2)*d = 0 m2/s2
900m2/s2 – 0 m2/s2 = (16.0m/s2)*d
900m2/s2 = (16.0m/s2)*d
(900m2/s2)/ (16.0m/s2) = d
(900m2/s2)/ (16.0m/s2) = d
56.3m = d
What is revealed is that the car skidded at a 56.3 meter distance. If you check, you will notice that this distance is about half of a soccer field which makes this a reasonable distance for skidding.
Before lifting off the ground, an airplane speeds up on a runway at 3.20 miles per second square. Before taking off, determine the traveled distance.
vi = 0 m/s
t= 32.8 s
a= +3.20 m/s2
0.5 * a * t + vi * t = d
0.5*( 3.20 m/s2) * (32. 8s)2+ (0 m/s) * (32. 8 s) =d
1720 m = d
At Great America, Jack Jones is riding the Giant Drop. How far will he fall and what will be his final velocity if he does a free fall of 2.60 seconds?
vi = 0 m/s
t = 2.6 s
a = -9.8 m
0.5*a*t2 + vi*t = d
0.5*(-9.8 m/s2)*(2.60 s) 2 + (0 m/s)*(2.60 s) = d
-33.1 m (- is direction indicated) = d
a*t + vi = vf
(2.60 s) * (-9.8 m/s2) + 0 = vf
-25.5 m/s (- is direction indicated) = vf
From parking, a car starts and uniformly accelerates over a distance of 110m over a time of 5.21 seconds. What is the car’s acceleration?
vi = 0 m/s
0.5*a*t2 + vi*t =d
0.5*(a)*(5.21 s)2 + (0 m/s)*(5.21 s)+ =110 m
(13.57 s2)*a = 110 m
(110 m)/ (13.57 s2) = a
8.10 m/ s2 = a
If a sled powered by rockets is used for testing how humans respond to acceleration, and in 1.83 seconds, accelerates to 444 m/s then what is the distance traveled by the sled and what is its acceleration?
t = 1.80s
vf = 44 m/s
vi = 0 m/s
vi*t + 0.5*a*t2 = d
(0 m/s)*(1.83 s) + 0.5*(243 m/s2)*(1.83 s)2 = d
0 m + 406 m = d
406 m = d
(Delta v)/t = a
(444 m/s – 0 m/s)/ (1.83 s) = a
243 m/s2 = a
Hope this helps! If you want more fundamentals about natural science and all the interesting subjects under its scope, here is a course entitled Quantum Physics: An Overview of a Weird World that introduces physics at a quantum level to beginners.