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1. describe the motion of the object in words. 2. calculate the velocit…

Question

  1. describe the motion of the object in words. 2. calculate the velocity of the object for each time the object changes its speed. (position vs. time graph: x - axis time (seconds) 0–8, y - axis position (meters) 0–4. graph line: from (0,0) to (2,4), horizontal to (3,4), then to (7,0), (8,0))

Explanation:

Response
Sub - question 1: Describe the motion of the object in words.
Brief Explanations
  • From \(t = 0\) to \(t = 2\) seconds: The object's position increases from 0 to 4 meters. Since position - time graph slope represents velocity, a positive slope here means the object is moving with a constant positive velocity (speeding up? No, constant velocity as slope is constant) in the positive direction.
  • From \(t = 2\) to \(t = 3\) seconds: The position of the object remains at 4 meters. A zero slope on a position - time graph means the object has zero velocity, so it is at rest.
  • From \(t = 3\) to \(t = 7\) seconds: The object's position decreases from 4 meters to 0 meters. A negative slope on the position - time graph means the object is moving with a constant negative velocity (moving in the negative direction) back towards the starting point.

Velocity \(v=\frac{\Delta x}{\Delta t}\), where \(\Delta x\) is the change in position and \(\Delta t\) is the change in time.

Step 1: Velocity from \(t = 0\) to \(t = 2\) s

\(\Delta x=4 - 0=4\) m, \(\Delta t = 2-0 = 2\) s.
\(v_1=\frac{\Delta x}{\Delta t}=\frac{4}{2}=2\) m/s.

Step 2: Velocity from \(t = 2\) to \(t = 3\) s

\(\Delta x = 4 - 4=0\) m, \(\Delta t=3 - 2 = 1\) s.
\(v_2=\frac{\Delta x}{\Delta t}=\frac{0}{1}=0\) m/s.

Step 3: Velocity from \(t = 3\) to \(t = 7\) s

\(\Delta x=0 - 4=- 4\) m, \(\Delta t = 7 - 3=4\) s.
\(v_3=\frac{\Delta x}{\Delta t}=\frac{-4}{4}=- 1\) m/s.

Answer:

From \(t = 0\) to \(t = 2\) s, the object moves with constant positive velocity (towards positive direction) from position 0 m to 4 m. From \(t = 2\) to \(t = 3\) s, the object is at rest (position remains 4 m). From \(t = 3\) to \(t = 7\) s, the object moves with constant negative velocity (towards negative direction) from 4 m back to 0 m.

Sub - question 2: Calculate the velocity of the object for each time the object changes its speed.