Question

An ant climbs to the very end of the second hand on a wall-mounted clock at exactly 9:15:00. The second hand is 13.5 cm long. When the time is exactly 9:17:30, (a) what distance did the ant travel? (b) What is the ant‟s displacement at this time? (c) What was the ant‟s average speed? (d) What is the ant‟s average velocity at this time? (e) At what time would the ant have traveled a distance of 100. M (give your answer to the nearest second).

Answer 1

Answer:

Explanation:

a) The second arm measures the minutes. The difference between 9:17:30 and 9:15:00 is 2 minutes 30 seconds. This means the second arm would have revolved 2.5 times.

The circumference = 2πr= 2π(13.5 cm) = 84.823 cm

Distance = 2.5 × 84.823 cm = 212 cm = 2.12 m

b) Displacement of an object is the shortest distance between the initial and final point. Since the second arm revolves 2 and half times, we use only the half times (a semicircle)

Displacement for semicircle = 2r = 2(13.5) = 27 cm

c) Average speed = distance / time

time = 2 minutes 30 seconds = 150 seconds

Average speed = 212 cm / 150 s = 1.41 cm/s

d) Average velocity = displacement / time = 27 cm / 150 s = 0.18 cm/s

e) number of revolutions = 100 m / circumference = 100 m / 0.848 m = 117.92

but 1 revolution = 1 minute

117.92 revolution = 117 minutes 55 seconds = 1 hour 57 minutes 55 seconds

Hence the time would be: 11:12:55