1 Škola: Gymnázium, Brno, Slovanské náměstí 7 Šablona: III/2 – Inovace a zkvalitnění výuky prostřednictvím ICT Název projektu: Inovace výuky na GSN prostřednictvím ICT Číslo projektu: CZ.1.07/1.5.00/ Autor: RNDr. Bc. Miroslav Hruška Tematická oblast: Kinematics of a particle (kinematika hmotného bodu) Název DUMu: Distance at uniformly accelerated motion – A Kód: VY_32_INOVACE_FY.2.13 Datum: Cílová skupina: Žáci středních škol s CLIL výukou fyziky v angličtině Klíčová slova:Acceleration, uniformly accelerated and decelerated motion, distance, deceleration, parabola Anotace: Prezentace slouží k zavedení vztahů pro dráhu rovnoměrně zrychleného a rovnoměrně zpomaleného pohybu. Prezentaci je vhodné použít při probírání nového učiva.

Kinematics of a particle 13. Distance at uniformly accelerated motion – A

3 Revision I  Decide which formula (see below) belongs to each type of rectilinear motion: 1. uniformly accelerated motion from rest 2. uniformly accelerated motion with an initial speed 3. uniformly decelerated motion

4 Revision II  Decide which formula (see below) belongs to each type of rectilinear motion: 1. uniformly accelerated motion from rest 2. uniformly accelerated motion with an initial speed 3. uniformly decelerated motion A) B) C)

5 Distance as a function of time I Graph of distance as a function of time (speeding up) at a = 2 m/s 2, v 0 = 0 m/s parabola

6 Distance as a function of time II Graph of distance as a function of time (speeding up) at a = 2 m/s 2, v 0 = 5 m/s parabola

7 Distance as a function of time III Graph of distance as a function of time (slowing down) at a = 2 m/s 2, v 0 = 10 m/s parabola

8  Uniformly accelerated motion from rest  Uniformly accelerated motion with an initial speed  Uniformly decelerated motion Formulas for distance

9  Speed (magnitude of velocity)  Distance  Slowing down  – in ±  Speeding up  + in ±  Speeding up from rest  v 0 = 0 m/s How to remember all formulas

10 Exercise I  A rocket reached a speed of 1.2 km/s in 20 seconds. Its movement was uniformly accelerated. Calculate the magnitude of the acceleration and the distance travelled during the given time interval.  v = 1.2 km/s = 1200 m/s  t = 20 s  v 0 = 0 m/s

11 Solution I  a = v ÷ t = 1200 m/s ÷ 20 s = 60 m/s 2  s = 0.5 × 60 × 20 2 m = m = 12 km  The rocket had an acceleration of 60 m/s 2, and the distance covered by it was 12 km.  v = 1200 m/s  t = 20 s  a, s = ?

12 Exercise II  Brakes of a car must ensure the car stops from a speed of 40 km/h in a maximum distance of 12.5 m on a dry road. Find the minimum magnitude of the deceleration and the time it takes.  v 0 = 40 km/h = 40 ÷ 3.6 m/s = 11.1 m/s  s = 12.5 m  v = 0 m/s

13  It is clever to let time run backwards!  v 0 = 0 m/s, v = 11.1 m/s  s = 12.5 m  a, t = ? Solution II ⇒ t = 2 × 12.5 m ÷ 11.1 m/s = 2.25 s ⇒ a = v ÷ t = 11.1 m/s ÷ 2.25 s = 4.9 m/s 2

That is all for now Thank you for your attention

15  Materiál je určen pro bezplatné používání pro potřeby výuky a vzdělávání na všech typech škol a školských zařízeních. Jakékoliv další využití podléhá autorskému zákonu. Zdroje:  Bednařík, M., Široká, M. Fyzika pro gymnázia, Mechanika. Dotisk 3. vyd. Praha: Prometheus, s. ISBN  Hornby, A. S. Oxford Advanced Learner’s Dictionary of Current English. 5. vyd. Oxford: Oxford University Press, s. ISBN