Veronika Kollár PreMed course

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1 Veronika Kollár PreMed course

2 The slope of a line y y y b y 1 x x 1 x The general equation of the line: f (x) = y = m x + b Where: b: intersection on the y axis m: the slope of the line x Intersection (b) - determines the point at which the line crosses the y-axis Slope (steepness) of the line: m y = x

3 Functions Linear and its slope The slope determines the raise of the linear (curve). tgα = opposite _ site adjacent _ site Positive Zero Negative y y y x x x y = x y = 1 y = -x

4 Kinematics-Dynamics Kinematics: doctrine of motion It investigates the parameters of body s motion (e.g. - displacement, time, velocity, acceleration). Dynamics: doctrine of forces It investigates effects of the forces on the body (e.g. - motion, deformation, interaction). It reveals the causes of motion.

5 KINEMATICS-Way, path, displacement displacement The way - the points in which the body is moving under the given time (the trajectory of motion). The path (or distance) - the section of the way. The displacement - the difference of the vectors which belong to two points of the path (describes the change in position of the object). - curve - scalar - vector

6 Average speed v = time of path the movement Average velocity v = displacement time of the movement Instantaneous speed v t = x t 5miles miles Speedaverage = = 5 0.hours hour

7 I. Uniform motion E.g.: Conveyor belt escalator Thebodyismoving inastraightlinepath. thedirectionofitsmovementdoesnotchange. it runs same lengths under the uniform time(constant speed). d const. v t = = d- displacement a=0 t- time v- velocity

8 Parameters of the uniform motion The velocity of the uniform motion : v = d t d t It is determined by two parameters: 1. starting place: d 0, (generally kept zero). velocity: v 0 0 d = d 0 + v t d (m) d 0 distance-time graph The dimensions of the velocity: 1 m s = 3 10 km 1/ 3600h = km h = 3.6 km h

9 Geometrically: the velocity is the slope of the length-time curve d(m) d d 0 d v = = t tgα distance travelled = time interval *velocity d

10 II. Constantly accelerated motion Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity. a v m = t s average E.g.: gravitational acceleration The acceleration of the motion is constant(the amplitude of its velocity changes uniformly).

11 Shape of the curve: parabola d(m) a v m = t s average a d = t distance-time graph second order or quadratic function

12 Constantly accelerated motion The path(distance): If the starting velocity v 0 = 0 m/s: d a = t if starting velocity v 0 0 m/s : The velocity: if starting velocity v 0 = 0 m/s: if starting velocity v 0 0 m/s: a d = v0 t + t v = a t v = v0 + a t Note! The area of the trapezoid (trapezium): T a + c = m

13 The slope of the velocity-time graph is constant! a = v v v v 1 = t t t 1 The area on the acceleration-time curve accords to the change of the velocity!

14 Calculation of the distance with the area on the velocity time plot. T a + c = m v + v 1 d = (t t 1) d (t t ) + t 1 3 = v

15 Exercises 1. A car is moving at first with a speed of 5 km/h for 4 minutes then 50 km/h for another 8 minutes. Finally it is moving 0 km/h which takes minutes. Calculate : a) the total travelled distance, in km. b) the average speed in SI units! a) 9 km b) 10,7 m/s. The airplane accelerates until 1 s before the take off (d=600 m). (a) Determine the acceleration. (b) Calculate the velocity at the end of 1 th s. (c) Determine the travelled distance in the last 1 s. a) 8,33 m/s b) 100 m/s c) 96 m

16 The experimental observation that all objects in free fall accelerate at the same rate, as noted by Galileo. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall.

18 Constantly accelerated motion. There are two important motion characteristics that are true of free-falling objects: Free-falling objects do not encounter air resistance. All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s. In a vacuum: F g =m*a

19 if starting velocity v 0 = 0 m/s: if starting velocity v 0 0 m/s: v = g t v = v0 + g t Distance: If the starting velocity v 0 = 0 m/s: if starting velocity v 0 0 m/s : g d = t g d = v0 t + t

20 3. Determine the distance traveled by the ball in the time domain and the position of the ball after the next times: 1.6 s,.4 s, 3. s. s: travelled distance d: the position the ball

21 4. The boy drops the ball from a roof of the house which takes 3 seconds to hit the ground. Calculate the velocity before the ball crashes to the ground. (g=10m/s²) v = g t m m v = 10 3s = 30 s s Let s found how height the ball has been dropped? We use 10 m/s² for g. Distance Traveled: 1 1 d = g t = 10 3 = 45m

22 5. How long does a body fall down from 10 cm, and how much is its velocity at the moment of the touch-down? (g=10 m/s ) t=0.14s v=1.4m/s

23 6. FromPécstoBudapest.Goingtheretakes3hours,back4 hours.(70 km) A. Whatis thespeedof trainback and forth? B. What is the mean speed for the whole trip? 90 and 67.5 km/h soo5 and m/s

24 7. A body is moving in the northeast direction with 1m/s. How greatis the northern and the eastern component of its velocity? v=0.7 m/s

25 8. John throws the ball straight upward and after 1 second it reaches its maximum height then it does free fall motion which takes seconds. Calculate the maximum height and velocity of the ball before it crashes the ground. (g=10m/s²)

26 9. A train is moving at first with a constant velocity (v = 36 km/h). After 15 min the train begins to accelerate and after 5 min it achieve 7 km/h and it does not accelerate any further. A.) How great the acceleration is (m/s )? After 5 min it starts to decelerate with m/s. B.) How long distance and how many seconds does the train need to stop? C.) How long is the total travelled distance? Calculate the average speed!

27 1) Vertically Launched Projectiles Free fall with initial velocity. Constantly accelerated motion ) Horizontally Launched Projectiles Uniform motion 3) Diagonally (0-90 )Launched Projectiles Javelin throw

28 The Pythagorean Theorem Starting point of the motion Acceleration: Velocity : Displacement: x direction a x = 0 y direction a y = g v = cosα v y vo sin gt x v o x vo (cosα) t = α = y = vo (sinα) t 1 gt

29 y Horizontal Vertical Forces No Yes, the force of gravity acts downward Acceleration No Yes "g" is downward at 9.8 m/s Velocity Constant Changing (by 9.8 m/s each second)

KINETICS: MOTION ON A STRAIGHT LINE. VELOCITY, ACCELERATION. FREELY FALLING BODIES

014.08.06. KINETICS: MOTION ON A STRAIGHT LINE. VELOCITY, ACCELERATION. FREELY FALLING BODIES www.biofizika.aok.pte.hu Premedical course 04.08.014. Fluids Kinematics Dynamics MECHANICS Velocity and acceleration