Question

A person walks steadily for 10.0 seconds away from the origin for a distance of 20.0 m, slows to a stop in 2.0 seconds while traveling another 2.0 m, turns around and speeds up for 2 seconds to a speed of 2.0 m/s in the other direction. the object goes back to the origin at a steady speed.

a) What was the acceleration from 10-12 seconds?
b) What was the acceleration from 12-14 seconds?
c) How long did the trip take?
d) What is the distance traveled for the trip?
e) What is the displacement of the trip?

Answer 1

The trip took 24.0 seconds and had a displacement of 0 meters.

Step-by-step explanation:

The trip can be divided into four parts:

1. Walking away from the origin: The person walks steadily for 10.0 seconds and covers a distance of 20.0 m.
2. Slowing down and traveling further: The person slows to a stop in 2.0 seconds while traveling another 2.0 m.
3. Turning around and speeding up: The person turns around and speeds up to a velocity of 2.0 m/s in the opposite direction in 2.0 seconds.
4. Returning back to the origin at a steady speed: This part takes the same amount of time as the first part, 10.0 seconds, since the person walks back at the same speed.

c) How long did the trip take?

The total time taken for the trip is the sum of the times for each part: 10.0 seconds + 2.0 seconds + 2.0 seconds + 10.0 seconds = 24.0 seconds

e) What is the displacement of the trip?

The displacement can be determined by calculating the difference between the total distances covered in the positive and negative directions. The total distance covered in the positive direction is 20.0 m + 2.0 m = 22.0 m. The total distance covered in the negative direction is also 22.0 m (since the person walks back at the same distance and speed). Therefore, the displacement of the trip is 0 meters.