Question

1. joey is walking to his neighbors house down the street.

a. how long does it take joey to reach his neighbors house?
b. how far away is joeys neighbors house?
c. how fast is joey walking to his neighbors house?

1. monica drives to rachels house, stays for a while, and then goes home. on her way home, her car breaks down.

a. how fast was monica driving on her way to rachels house?
b. how long did she spend at rachels?
c. how far away was monica from her house when her car broke down?
d. what was the total distance monica had traveled when her car died?

Explanation:

Step 1: Analyze Joey's problem

• a. From the position - time graph for Joey, the time taken to reach the neighbor's house is the time value at the end - point of the motion. The graph ends at \(t = 15\ s\), so the time taken is \(15\ s\).
• b. The distance to the neighbor's house is the final position value. From the graph, the final position \(x = 30\ m\).
• c. The speed \(v\) is given by the slope of the position - time graph. The slope \(m=\frac{\Delta x}{\Delta t}\), where \(\Delta x=30 - 0=30\ m\) and \(\Delta t = 15-0 = 15\ s\). So \(v=\frac{30}{15}=2\ m/s\).

Step 2: Analyze Monica's problem

• a. For Monica's motion to Rachel's house, the speed is the slope of the first non - horizontal part of the graph. The position changes from \(x = 0\) to \(x = 80\ m\) in time \(t = 1\ h\). So the speed \(v=\frac{80 - 0}{1}=80\ m/h\).
• b. The time spent at Rachel's is the time interval when the position is constant. The position is constant from \(t = 1\ h\) to \(t = 2\ h\), so the time spent is \(2 - 1=1\ h\).
• c. When the car breaks down, the position is \(40\ m\) (from the graph where the slope changes to zero on the return journey).
• d. The total distance traveled when the car breaks down: She travels \(80\ m\) to Rachel's and then \(40\ m\) back. So the total distance \(d=80 + 40=120\ m\).

Answer:

1.

• a. \(15\ s\)
• b. \(30\ m\)
• c. \(2\ m/s\)

2.

• a. \(80\ m/h\)
• b. \(1\ h\)
• c. \(40\ m\)
• d. \(120\ m\)