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ZveřejnilVratislav Král
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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.1.07/1.5.00/34.0940 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:1. 8. 2013 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.
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Kinematics of a particle 13. Distance at uniformly accelerated motion – A
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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
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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)
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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
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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
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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
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8 Uniformly accelerated motion from rest Uniformly accelerated motion with an initial speed Uniformly decelerated motion Formulas for distance
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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
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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
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11 Solution I a = v ÷ t = 1200 m/s ÷ 20 s = 60 m/s 2 s = 0.5 × 60 × 20 2 m = 12 000 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 = ?
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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
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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
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That is all for now Thank you for your attention
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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, 2004. 288 s. ISBN 80- 7196-176-0 Hornby, A. S. Oxford Advanced Learner’s Dictionary of Current English. 5. vyd. Oxford: Oxford University Press, 1995. 1428 s. ISBN 0-19-431423-5
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