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17 December, 07:33

Mary walked north from her home to sheila's home, which is 4.0 kilometers away. then she turned right and walked another 3.0 kilometers to the supermarket, which is 5.0 kilometers from her own home. she walked the total distance in 1.5 hours. what were her average speed and average velocity?

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  1. 17 December, 07:57
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    To find average speed, we can use the formula:

    speed (m/s) = total distance (m) / time (s)

    total distance = 4km + 3km = 7km = 7000m

    time = 1.5 hours = 5400s

    speed = 7000/5400 = 1.30m/s

    To find average velocity, we can use the same formula, but replacing total distance with total displacement.

    total displacement = 5km = 5000m (if we use the distance to Sheila's house and the distance from her house to the supermarket as the vector arrows for vector addition, we get 5km, as stated in the question)

    velocity = 5000/5400 = 0.926m/s

    Because this is a velocity, we need the direction. To find this we can use the formula:

    tan (theta) = opposite / adjacent

    tan (theta) = 3000/4000

    sin (tan (theta)) ^-1 = sin (3000/4000) ^-1

    theta = 48.6° East of North or 41.4° North of East

    So the average velocity is 0.926m/s [48.6° East of North / 41.4° North of East]
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