physics Joseph Jogs from one end A to the other end B of a straight 300m road in 2 minutes 30 second then turns around and jogs

Question

physics
Joseph Jogs from one end A to the other end B of a straight
300m road in 2 minutes 30 second then turns around and jogs 100 m back to point C in another minute. What are Josef’s Average speeds and velocities in jogging from
a) A to B
and
b) A to C

pls give answer with explination do not send link​

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Orla Orla 4 years 2021-08-27T10:50:24+00:00 1 Answers 18 views 0

Answers ( )

  1. binhan
    0
    2021-08-27T10:52:22+00:00
    Reply

    Answer:

    a) Josef jogging from A to B;

    i. speed = 2 m/s

    ii. velocity = 2m/s

    a) a) Josef jogging from A to C;

    speed = 1.9 m/s

    ii. velocity = 2.2 m/s

    Step-by-step explanation:

    One major difference between speed and velocity is that speed is a scalar quantity, while velocity is a vector quantity.

    a) Josef jogging from A to B;

    distance = 300 m

    time = 2.5 seconds = 150 seconds

    i. his average speed = \frac{distance}{time}

                                   = \frac{300}{150}

                                   = 2 m/s

    ii. his average velocity = \frac{displacement}{time}

                                   = \frac{300}{150}

                                   = 2 m/s

    b) Josef jogging from A to C,

    distance = 300 + 100 = 400 m

    time = 2.5 + 1 = 3.5 minutes

            = 210 seconds

    i. his average speed = \frac{distance}{time}

                                      = \frac{400}{210}

                                      = 1.9 m/s

    ii. his average velocity = \frac{\sqrt{V_{1} ^{2} + V_{2} ^{2} } }{2}

    V_{1} ^{2} = 4 m/s

    V_{2} ^{2} = 1.9 m/s

    his average velocity = \frac{\sqrt{4^{2} + 1.9^{2} } }{2}

                                      = 2.2 m/s

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