Question

1)A student observes an ant travelling at 1 cm/s. At the time of 1:02:32 pm the ant is at the 5.0 cm position on the meter stick. Where on the meter stick will the ant be at 1:03:10 pm?

Answer 1

43.0 cm

Step-by-step explanation:

The ant is traveling at a speed of 1 cm/s. At 1:02:32 pm, the ant is at the 5.0 cm position on the meter stick. The time difference between 1:02:32 pm and 1:03:10 pm is 38 seconds. Since the ant is traveling at a speed of 1 cm/s, in 38 seconds it will have traveled a distance of 38 cm. Therefore, at 1:03:10 pm, the ant will be at the position of 5.0 + 38 = 43.0 cm on the meter stick.

Answer 2

The time interval between 1:02:32 pm and 1:03:10 pm is 38 seconds. During this time interval, the ant travels at a speed of 1 cm/s, so it will travel a distance of:

distance = speed x time

distance = 1 cm/s x 38 s

distance = 38 cm

Starting at the 5.0 cm position on the meter stick, the ant will travel 38 cm in 38 seconds, ending up at:

5.0 cm + 38 cm = 43.0 cm

Therefore, the ant will be at the 43.0 cm position on the meter stick at 1:03:10 pm.