Question

a speedboat travels from the dock to the first buoy, a distance of 20 meters, in 18 seconds. it began the trip at a speed of 0 meters per second. by the time it reached the buoy, it was going 10 meters per second.
what was the average velocity of the boat?
-3.33 m/s²
-1.11 m/s
+1.11 m/s
+3.33 m/s

Explanation:

Step1: Recall the formula for average velocity

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

Step2: Identify the values of \(\Delta x\) and \(\Delta t\)

The displacement \(\Delta x = 20\) meters (from the dock to the buoy) and the time interval \(\Delta t=18\) seconds.

Step3: Calculate the average velocity

\(v_{avg}=\frac{20}{18}\approx1.11\) m/s. The direction is positive as the boat is moving from the dock to the buoy (assuming the direction from dock to buoy is the positive direction).

Answer:

\(+ 1.11\) m/s (the third option)