Question

question 5 (3 points) (01.10 mc) a position versus time graph is shown below. position vs. time graph (x - axis: time (s), y - axis: position (m, north)) use the graph to determine the average velocity in m/s of the object at the end of 2 seconds. options: a) 1 m/s north, b) 2 m/s north, c) 4 m/s north, d) 5 m/s north

Explanation:

Step1: Recall average velocity formula

Average velocity \( v_{avg} = \frac{\Delta x}{\Delta t} \), where \( \Delta x \) is displacement and \( \Delta t \) is time interval.

Step2: Determine displacement and time

At \( t = 0 \) s, position \( x_0 = 0 \) m. At \( t = 2 \) s, from the graph, position \( x_2 = 4 \) m (assuming the y - axis grid: first non - zero point at t = 1 is 4 m, and it stays constant till t = 2). \( \Delta x=x_2 - x_0=4 - 0 = 4 \) m, \( \Delta t = 2-0 = 2 \) s.

Step3: Calculate average velocity

\( v_{avg}=\frac{\Delta x}{\Delta t}=\frac{4}{2}=2 \) m/s north. Wait, wait, no, wait. Wait, at t = 1s, position is 4m? Wait, maybe I misread. Wait, the first segment: from t = 0 to t = 1, position goes from 0 to 4? Then from t = 1 to t = 2, it's constant. So at t = 2s, position is 4m. So \( \Delta x = 4 - 0=4 \) m, \( \Delta t = 2 \) s. Wait, no, \( v=\frac{4}{2}=2 \)? Wait, but let's check again. Wait, the formula is \( v=\frac{\text{displacement}}{\text{time}} \). Displacement is change in position. At t = 0, x = 0. At t = 2, x = 4. So \( \Delta x = 4 \) m, \( \Delta t = 2 \) s. So \( v=\frac{4}{2}=2 \) m/s north. So the correct option is b.

Answer:

b. 2 m/s north