Kinematics- What is it?
Kinematics is defined as the branch of mechanics that involves the motion of objects without involving the forces that causes the motion. In our study of kinematics, we did a series of labs and activities that led us to derive 3 main equations. These 3 equations, which will be further detailed below, involve 6 different variables involving velocity, position, time, and acceleration. We made use of technology using programs such as LoggerPro, and it's motion detector or video analysis feature.
Terminology
Position= the place where someone or something is in relation to something else (initial position: where object starts, final position: where object finishes). Represented by variable x
Distance= the total amount someone or something travels in a given time. Distance is always positive.
Displacement= the difference between the final and initial position, or change in x. Not the same as distance, represented by the variable Δx. Displacement can be negative or positive.
Distance= the total amount someone or something travels in a given time. Distance is always positive.
Displacement= the difference between the final and initial position, or change in x. Not the same as distance, represented by the variable Δx. Displacement can be negative or positive.
Position time graphs
Interpreting Velocity from position time graphs |
A position time graph shows the position of an object at different points in time. For example, if I were to run a mile, starting at 0meters, a measurement of my position would be taken at each minute (t), and points (t,x) would be plotted. As seen in another example on the left, as time (t in seconds) goes on, position (x in meters) increases. As seen by the linear model that was fit to the data points, the slope is the same, meaning that for every increase in time, there is the same amount of increase in position. The slope in this case is 3, so for every 1 second increase, the position increases by 3 meters. The y intercept here is 0, meaning that when t=0, position is also 0.
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The slope of a velocity time graph, as described above, is the change of position over the change in time. This is also known as speed (m/s). However, speed simply has a magnitude and doesn't have a direction. If we know the direction of the movement of an object, the change in position over the change in time is known as the velocity (m/s). Therefore, the slope of a position time graph is the velocity. The velocity at any point is called instantaneous velocity, regardless of type of position-time model. In the example on the right, our position time graph is increasing at a decreasing rate. Therefore, our velocity is becoming more negative as time goes on, but it is still moving in a positive direction. The velocity time graph reflects that: velocity becomes more negative. It is decreasing at a constant rate which is why the velocity time graph is a linear model.
Interpreting velocity time graphsWhere the velocity time graph has a positive slope implies that there is a positive acceleration- the rate at which the object is speeding up is increasing. Where the slope is 0, or where there is a horizontal line, indicates that the velocity of the object is constant, it is not speeding up or slowing down. Where the model has a negative slope indicates a negative acceleration (deacceleration), indicating that the object is slowing down.
Determining acceleration and starting velocity from position time graphAcceleration of a position time graph can be determined by finding the initial position and velocity of an object. If the slope of a position time graph is constant, meaning a linear model, the velocity would be a horizontal line, and therefore the slope of that (or acceleration) would be 0. If the rate of change of a position time graph is not constant, then the velocity would not be 0. To determine the acceleration of a nonlinear curve, we would use the kinematic equation: Δx=1/2at^2+v0t.
Further, positive acceleration has a position time graph opening upward and vice versa. |
Determining position, displacement, and distance
From just looking at a velocity time graph, we can also determine position, distance, and displacement. Find the area under the curve to find displacement. To find distance, subtract the area under the curve that is below the x axis from the area under the curve above the x axis. For example, in the figure to the right, the displacement would be the sum of the area of trapezoid with length PQ plus the area of the trapezoid with length RS. To find the displacement, substract trapezoid RS from trapezoid PQ. If there were enough mathematical information provided, you could use the three kinematic equations to determine xi, xf, or Δx.
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Connecting representations of motions
Position Time Graph--> Velocity Time Graph--> Acceleration Time Graph
- graphs come from the slope of the previous graph
- slope of position time graph is velocity, slope of velocity time graph is acceleration
- graphs come from the slope of the previous graph
- slope of position time graph is velocity, slope of velocity time graph is acceleration
Acceleration can be determine by a graphing utility such as LoggerPro using a strobe diagram. A strobe diagram "uses dots to represent the position and time of an object each second." If the object is accelerating the distance between the dots should increase as time increases. If the object is moving at a constant speed and uniform accelerating, then the distance between the dots should remain the same. For example, when you drop a ball, the speed of the ball picks up (accelerates) as time goes on. See motion of ball down a ramp below, and see how the distance between the ball at different points of time is increasing as time goes on.
3 main kinematic equations
1) xf-xi=1/2at^2+vit
2) vf=at+vi
3) vf^2=vi^2+2a(xf-xi)
2) vf=at+vi
3) vf^2=vi^2+2a(xf-xi)
problem solving
Most of the problems that involve kinematics are solvable by manipulating these 3 equations and the 6 variables within. The six variables are: position final, position initial, acceleration, change in time, velocity initial, and velocity final. Each problem won't give 5 variables and make you solve for one- they may give you one or two, and then based on subtle context information or such, figure out one or two. Though it may take a while to get comfortable using these equations, practicing several different types of problems will help solidify the ideas. Some examples:
1) A cyclist is initially traveling at 10 m/s, they accelerate at 2 m/s2 for 3 seconds. What is their velocity after the acceleration? How much distance did they cover during the acceleration? (KINEMATICS)
- First, identify the variables we are given.
initial velocity= 10m/s, a= 2m/s^2, t=3
- We can choose one of the three equations to plug our variables into
vf = vi + a t---> vf=10+2(3)= 16 m/s
- Now that we've solved for the velocity after the acceleration, we can use these 4 known variables to figure out the distance covered during the acceleration.
d = vi(t) + ½ a t2--> d=10(3)+1/2(2)(3)^2= 39m
2) A race car reaches a velocity of 85 m/s after traveling 200 m from the starting line. Assuming the car keeps that same velocity, how far from the starting line would the car be after 15 seconds? (CONSTANT VELOCITY)
- Determine given information. Use equations to solve for unknown variable.
vf=85m/s, t=15s, xi= 200m. If car keeps same velocity, acceleration must be 0 (a=0)
xf=1/2at^2+vi(t)+xi---> xf=vi(t)+xi= 85(15)+200= 1475m
3) With an initial velocity of 20 km/h, a car accelerated at 8 m/s2 for 10 seconds. (UNIFORM ACCLERATION)
a) What is the position of the car at the end of the 10 seconds?
b) What is the velocity of the car at the end of the 10 seconds?
- Identify known variables. (vi=20, a=8, t=10) .... 20km/hr=5.6m/s
x=1/2at^2+vit---> 1/2(8)(10)^2+5.6(10)= 456m
- Using these variables, we can find the velocity of the car at the end of 10 seconds by using a different kinematic equation.
vf=at+vi---> 8(10)+5.6= 85.6m/s
1) A cyclist is initially traveling at 10 m/s, they accelerate at 2 m/s2 for 3 seconds. What is their velocity after the acceleration? How much distance did they cover during the acceleration? (KINEMATICS)
- First, identify the variables we are given.
initial velocity= 10m/s, a= 2m/s^2, t=3
- We can choose one of the three equations to plug our variables into
vf = vi + a t---> vf=10+2(3)= 16 m/s
- Now that we've solved for the velocity after the acceleration, we can use these 4 known variables to figure out the distance covered during the acceleration.
d = vi(t) + ½ a t2--> d=10(3)+1/2(2)(3)^2= 39m
2) A race car reaches a velocity of 85 m/s after traveling 200 m from the starting line. Assuming the car keeps that same velocity, how far from the starting line would the car be after 15 seconds? (CONSTANT VELOCITY)
- Determine given information. Use equations to solve for unknown variable.
vf=85m/s, t=15s, xi= 200m. If car keeps same velocity, acceleration must be 0 (a=0)
xf=1/2at^2+vi(t)+xi---> xf=vi(t)+xi= 85(15)+200= 1475m
3) With an initial velocity of 20 km/h, a car accelerated at 8 m/s2 for 10 seconds. (UNIFORM ACCLERATION)
a) What is the position of the car at the end of the 10 seconds?
b) What is the velocity of the car at the end of the 10 seconds?
- Identify known variables. (vi=20, a=8, t=10) .... 20km/hr=5.6m/s
x=1/2at^2+vit---> 1/2(8)(10)^2+5.6(10)= 456m
- Using these variables, we can find the velocity of the car at the end of 10 seconds by using a different kinematic equation.
vf=at+vi---> 8(10)+5.6= 85.6m/s
Projectile motion
Projectile= an object that is in the air but is not touching the air
- projectile motion has both vertical AND horizontal motion
X motion, or horizontal motion, is how far the object travels and has a constant velocity. Y motion, or vertical motion is how long the object is in the air. The derived equation for the distance a projectile moves is: d=sqrt(2arl)(2h/g)
- projectile motion has both vertical AND horizontal motion
X motion, or horizontal motion, is how far the object travels and has a constant velocity. Y motion, or vertical motion is how long the object is in the air. The derived equation for the distance a projectile moves is: d=sqrt(2arl)(2h/g)
EXtra Resources
https://www.youtube.com/watch?v=8NLzuURxFwY- PROJECTILE MOTION VIDEO
https://www.crackap.com/ap/physics-1/test20.html - KINEMATICS PROBLEMS https://www.youtube.com/watch?v=nUb7xfkc0Ac - POSITION, VELOCITY, ACCLERATION TIME GRAPH VIDEO https://www.physicsclassroom.com/class/1DKin/Lesson-1/Distance-and-Displacement- DISTANCE VS DISPLACEMENT |