IB Physics SL

By Kimberly Miller,2014-07-02 14:48
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IB Physics SL

IB Physics SL GOHS

2.1 Worksheet: Displacement

    xxxConcept Questions fi2. If the velocity of a particle is zero, can the particle’s

    acceleration be zero? Explain.

    4. The speed of sound in air is 331 m/s. During the next thunderstorm, try to estimate your distance from a lightning bolt by measuring the time lag between the flash and the thunderclap. You can ignore the time it takes for the light flash to reach you. Why?


    3. Two boats start together and race across a 60-km-wide lake and back. Boat A goes across at 60 km/h and returns at 60 km/h. Boat B goes across at 30 km/h, and its crew, realizing how far behind it is getting, returns at 90 km/h. Turnaround times are negligible, and the boat that completes the round trip first wins. (a) Which boat wins and by how much? (Or is it a tie?) (b) What is the average velocity of the winning boat?

    4. The Olympic record for the marathon is 2 h, 9 min, 21 s. The marathon distance is 26 mi, 385 yd. Determine the average speed (in miles per hour) of the record.

6. A graph of position versus time for a certain

    particle moving along the x-axis is shown in Figure

    P2.6. Find the average velocity in the time intervals

    from (a) 0 to 2.00 s, (b) 0 to 4.00 s, (c) 2.00 s to 4.00

    s, (d) 4.00 s to 7.00 s, and (e) 0 to 8.00 s.


    Concept Questions

     2. Yes. Zero velocity means that the object is at rest. If the object also has zero

    acceleration, the velocity is not changing and the object will continue to be at rest.

     4. You can ignore the time for the lightning to reach you because light travels at the

    8310 msspeed of , a speed so fast that in our day-to-day activities it is

    essentially infinite.


    2.3 (a) Boat A requires 1.0 h to cross the lake and 1.0 h to return, total time 2.0 h.

    Boat B requires 2.0 h to cross the lake at which time the race is over.

    Boat A wins, being 60 km ahead of B when the race ends.

     (b) Average velocity is the net displacement of the boat divided by the total

    elapsed time. The winning boat is back where it started, its displacement

    zerothus being zero, yielding an average velocity of .

    x2.4 The average speed is where vavt

    (:3 ft1 mi(: x26.0 mi385 yard26.2 mi!,?,;;;;;;1 yard5280 ft,~,~

    1 h1 h(:(:and t!,?,2.00 h+9.00 min21.0 s2.16 h;;;;;;;;60 min3600 s,~,~

    26.2 mimiThus, v,,12.2 av2.16 hh

    !?x10.0 m05.00 ms2.6 (a) ,,,vav!?2.00 s0t

    !?x5.00 m01.25 ms (b) ,,,vav!?4.00 s0t

    !?x5.00 m10.0 m2.50 ms (c) ,,,vav!?4.00 s2.00 st

    ??5.00 m5.00 mx3.33 ms (d) ,,,vav!?7.00 s4.00 st

xx!?x00210 (e) ,,,,vav!??8.00 s0ttt21

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