Question

a polar bear starts at the north pole. it travels 1.0 km south, then 1.0 km east, then 1.0 km north, then 1.0 km west to return to its starting point. this trip takes 45 min. what was the bears average velocity? 0 km/h 0.09 km/h 4.5 km/h 5.3 km/h question 6 (1 point) if you run a complete loop around an outdoor track (400 m), in 100 s, your average velocity is 0.25 m/s. 4.0 m/s. 40,000 m/s. zero. question 7 (1 point) if the position versus time graph of an object is a vertical line, the object is moving with constant non - zero speed. moving with constant non - zero acceleration. at rest. moving with infinite speed.

Explanation:

Question (About Polar Bear's Average Velocity)

Step1: Recall average velocity formula

Average velocity is defined as the displacement divided by time, \( v_{avg} = \frac{\Delta x}{t} \), where \( \Delta x \) is the displacement (change in position) and \( t \) is the time taken.

Step2: Determine displacement

The polar bear starts at the North Pole and returns to the starting point. So, the displacement \( \Delta x = 0 \) km (since initial and final positions are the same).

Step3: Calculate average velocity

Given time \( t = 45 \) min \( = \frac{45}{60} = 0.75 \) h. Using the formula \( v_{avg} = \frac{\Delta x}{t} = \frac{0\ \text{km}}{0.75\ \text{h}} = 0\ \text{km/h} \).

Step1: Recall average velocity formula

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

Step2: Determine displacement

Running a complete loop means starting and ending at the same point, so displacement \( \Delta x = 0 \) m.

Step3: Calculate average velocity

Time \( t = 100 \) s. Then \( v_{avg} = \frac{0\ \text{m}}{100\ \text{s}} = 0\ \text{m/s} \) (or "zero" as an option).

Step1: Analyze position - time graph

In a position - time graph, the slope represents velocity (\( v=\frac{\Delta x}{\Delta t} \)). A vertical line means that for a change in time (\( \Delta t>0 \)), the change in position (\( \Delta x = 0 \))? No, wait, a vertical line would imply that at a single instant of time, the position is changing infinitely (since \( \Delta t = 0 \) for a vertical line, and \( v=\frac{\Delta x}{\Delta t} \), if \( \Delta t = 0 \), velocity is infinite). Let's think again: If the graph is vertical, it means that at the same time, the object is at different positions, which would imply that the time is not changing but position is, so \( \Delta t = 0 \) and \( \Delta x
eq0 \), so \( v=\frac{\Delta x}{\Delta t} \) would be infinite (since \( \Delta t = 0 \)). So the object is moving with infinite speed.
Let's check the options:

• "moving with constant non - zero speed": For a constant non - zero speed, the graph would be a straight line with non - zero slope (not vertical).
• "moving with constant non - zero acceleration": Acceleration is related to the slope of the velocity - time graph, not position - time graph directly, and a vertical position - time graph doesn't imply constant non - zero acceleration.
• "at rest": If at rest, the position - time graph is a horizontal line (slope = 0), not vertical.
• "moving with infinite speed": Since \( v=\frac{\Delta x}{\Delta t} \), and for a vertical line \( \Delta t = 0 \) (same time, different positions), so \( v \) is infinite.

Answer:

0 km/h

Question 6 (Running Around Track)