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A cyclist at a grocery store, 180m from his home decides to go to the post office which is 510m from his home in the same direction. He takes 45 seconds to make the trip. He then returns home, taking a further 60 seconds.
a. What was his total displacement?
b. What was his average velocity in going from the grocery store to the post office?
c. How much was his average speed in making the entire trip

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Onyebuchinnajiidn

(a) The total displacement is zero

(b) The average velocity from the grocery store to the post office is 11 m/s.

(c) The average speed of the entire trip is 9.71 m/s

The given parameters;

  • initial position = 180 m at store
  • final position = 330 m at post office
  • time for the forward trip = 45 s
  • time for the backward trip = 60 s

(a) The total displacement is the change in his position;

Displacement = forward position - backward position

Displacement = 510 m - 510 m = 0

(b) The average velocity from the grocery store to the post office;

Displacement = 510 m - 180 m = 330 m

Time of motion from his home to grocery store, = (60 - 45) = 15 s

Time of motion to post office = 45 s - 15 s = 30 s

[tex]average \ velocity = \frac{330}{30} = 11 \ m/s[/tex]

(c) The average speed of the entire trip;

total distance = 510 m forward + 510 m backward = 1020 m

total time = 45 s + 60 s = 105 s

[tex]average \ speed = \frac{1020}{105} \\\\average \ speed = 9.71 \ m/s[/tex]

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