Go to the Moving Man PhET simulation https://phet.colorado.edu/sims/cheerpj/moving-man/latest/moving-man.html?simulation=moving-man to chart. Set all settings to zero by using “Reset All”. Set position to x = -10 m, velocity to 2 m/s, and acceleration to 0.5 m/s². Check the Record button and run the simulation. Stop when the man is approximately at 8 m position. Without resetting the values, now change the velocity to -2 m/s and acceleration to -0.5 m/s². Run the simulation. Stop when the man is approximately at x = -10 m position. Question 1: Explaining whether acceleration is constant or varied? Did the man speed up during his movement? If so, give the initial and final position values when he sped up. Explain your reasoning. Did the man slow down during his movement? If so, give the initial and final position values when he slowed down. Explain your reasoning.
The Correct Answer and Explanation is:
In the first part of the simulation, the man starts at x=−10 mx = -10 \, \text{m}x=−10m, with a velocity of 2 m/s2 \, \text{m/s}2m/s and an acceleration of 0.5 m/s20.5 \, \text{m/s}^20.5m/s2. Since acceleration is constant and positive, the velocity of the man will increase as he moves forward, meaning he speeds up. The position will continuously increase as long as acceleration remains constant. When the man reaches approximately x=8 mx = 8 \, \text{m}x=8m, his velocity has increased from 2 m/s2 \, \text{m/s}2m/s to a higher value due to the constant positive acceleration.
For the second part of the simulation, the velocity is changed to −2 m/s-2 \, \text{m/s}−2m/s, and the acceleration is set to −0.5 m/s2-0.5 \, \text{m/s}^2−0.5m/s2. Here, the negative velocity indicates that the man is moving in the opposite direction, and the negative acceleration means that the man is slowing down. As he moves back towards x=−10 mx = -10 \, \text{m}x=−10m, the velocity becomes less negative (i.e., the man is slowing down), and the position will move closer to −10 m-10 \, \text{m}−10m.
Regarding the acceleration:
- Acceleration is constant in both parts of the motion since it does not change over time. It is only the direction of acceleration that switches from positive (speeding up) to negative (slowing down) when the direction of motion changes.
- The man sped up in the first phase when moving forward because the velocity increased due to the positive acceleration. His initial position was −10 m-10 \, \text{m}−10m, and his final position when he sped up was near 8 m8 \, \text{m}8m.
- The man slowed down in the second phase as he moved in the opposite direction. The initial position was near 8 m8 \, \text{m}8m, and he slowed down until he reached about −10 m-10 \, \text{m}−10m.
In summary, the man sped up during the forward motion and slowed down when moving back, with constant acceleration throughout.
