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Basics of Physics Course I: Kinematics

Matthias Thelen 1

Last updated on 25. May 2020

This chapter gives you an overview of all the basics of kinematics that are important in biomechanics. The aim is to give you an introduction to the physics of biomechanics and maybe to awaken your fascination for thereof. The topics movement (location, velocity, acceleration) are covered. In order to illustrate the physical laws in a practical way, there is one example from physics and one from biomechanics for each topic area. Do not be surprised if you see a \Delta (a so-called delta) in front of some variables. This indicates that there is a change in some way, such as a change in speed or time. We hope it does not confuse too much, but it is important to us that all of our formulas are mathematically correct. Have fun! 

 

  1. Movement

  2. Location

  3. Velocity

  4. Acceleration

 

1. Movement

The analysis of movement plays a central role in biomechanics. The physical quantities, the location, the speed and the acceleration are used to describe the movement.

2. Location

In physics, the location s(t) indicates the position of a body in space at a certain point time t . Its unit is meter [m] .

A race car is located 2 seconds after the start (i.e. t = 2s ) on a straight race track 100 meters from the start. So the location of our race car would be at 100 meters if we look from the start. As you can see, a reference point is always needed to determine the location. In our case this is the start of the track. From a mathematical point of view, the reference point is the origin of the coordinate system and this example is a one-dimensional case, because the location of the race car can be determined with only one distance to the reference point. 

An example of a three-dimensional case would be the location of an airplane in relation to its landing airport. For example, the plane is still 40km to the east, 20km to the north and at an altitude of 10km. So the location of the plane would be (4000m/2000m/1000m). As you can see, besides a reference point you always need fixed axes (here east, north, altitude), along which the position of the body (e.g. airplane, race car) can be determined.  

In beach volleyball, a player makes an attacking shot above the edge of the net. The aim is to determine where on the pitch she hits the ball. As shown in the two examples of physics, we must first determine a reference point, i.e. the origin of our coordinate system and the axes of the coordinate system. The reference point is here one of the two points where the center line and the outline come together. The first axis is the center line, the second axis is the outline and the third axis of the coordinate system is the vertical, i.e. the height. So we can determine that the player hits the ball 0.5m from the outline, 0.1m from the net and at a height of 2.4m. So the location of the hand would be (0.5m/0.1m/2.4m). 

 

3. Velocity

In physics, the velocity v indicates how fast a body moves and thus changes its location. The unit is meter per second [\frac{m}{s}] . The velocity   v  is the mathematical derivation of the location function according to time t and is therefore described by the general formula v = \frac{\Delta s}{\Delta t} , where \Delta s is the change of position in the a certain period of time \Delta t .

Let's take a look at the race car again. The race car is not only 100 meters away from the start but also has a certain speed of v = 80 \frac{m}{s}  . This means that the race car drives 80 meters in 1 second. So, if we wait another 1 second, it will not be 100 meters from the start but 180 meters. 

So, the formula for the location at constant velocity is

s(t) = v ∗ t + s_0

s(1s) = 80 \frac{m}{s} ∗ 1s + 100m = 180m

Here s(t) is the location where the car is after time t , t  is the time the car is driving with speed v and s_0  is the starting location. 

If we stay with beach volleyball, we are interested in the speed of the hand just before the attack stroke, for example. If the hand and the whole arm of the beach volleyball player moves e.g. with a speed of v = 2 \frac{m}{s} , this indicates with which speed the ball will move after the stroke. You will find out exactly how this is connected in the Impulse section in the Basics of Physics Course II: Kinetics.  

4. Acceleration

In physics, the acceleration a indicates how the velocity of a body changes. Its unit is meter per second squared [\frac{m}{s^2}] . The acceleration a is the mathematical derivative of the velocity function with respect to time and thus the second derivative of the location function. This is described by the general formula a = \frac{\Delta v}{\Delta t} , where \Delta v is the change of velocity in the a certain period of time \Delta t .

The race car has an initial speed v_0 = 0\frac{m}{s} at the start before it starts to move, because it is not moving yet. After 100 meters it then has a speed of v = 80\frac{m}{s} and it needs 2 seconds (so t = 2s ) to change its speed from 0\frac{m}{s} to 80\frac{m}{s} . To do this the car has to accelerate. So the car has become 40\frac{m}{s} faster every second. So the acceleration of the car was a =  40\frac{m}{s^2} . The second is squared during acceleration, because it indicates the speed "per" second, i.e. \frac{v}{t}   So the formula for the speed at constant acceleration is 

v(t) = a ∗ t + v_0

v(2s) = 40\frac{m}{s^2} ∗ 2s + 0\frac{m}{s} = 80\frac{m}{s}  

Here v(t) is the speed of the car after the time t, t is the time the car is accelerated with the acceleration a and v_0  is the initial speed of the car.  

When swinging out for the attack stroke in beach volleyball, the arm and hand have a speed of v_0 = 0\frac{m}{s} at the beginning. But since the hand and arm have a velocity of v_1 = 2\frac{m}{s} at the end of the stroke when hitting the ball, the hand and arm were accelerated by the muscles of the player. It is estimated that half a second passes from the back swing to the impact. So the hand and arm are accelerated from v_0 = 0\frac{m}{s} to v_1 = 2\frac{m}{s} in \Delta t = \frac{1}{2}s . This corresponds to change of velocity of \Delta v = 2\frac{m}{s} and thus to an acceleration of a = 4\frac{m}{s^2}

  1. andres psijas andres psijas

    that information was great, it´s important understand the physics but also a specific one for the biomecanichs

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