Speedboat Velocity

Question

A speedboat moves west at 108 km/h for 26.3 min. It then moves at 60.0° south of west at 90.0 km/h for 10.0 min.

What is the average speed for the trip?

What is the magnitude of average velocity for the trip?



Answer 1

The trick to this one is that when the boat changed velocity it changed its acceleration.

It went at a constant velocity for a while (acceleration of 0),

it accelerated at some value to change velocity, speed and direction,

then it traveled at a constant velocity (acceleration of 0) again.

The acceleration changed from zero to something to zero again.

** When the acceleration is not constant the average velocity is NOT the average of the two velocities.

Also, the only equations that work are the definitions of speed, velocity, and acceleration.

All the equations that are derived for those definitions using v(avg) = (v1 + v2)/2 do not work.

SO, use the definition of speed --- speed = distance / time --- to find how far it went at each speed.

Then you can find the total distance by adding them and divide by time to find the average speed.

Velocity is a vector quantity so you cannot just add up the two distances.

Use vector addition.

You will need to add the two displacement vectors to find the resultant change in position.

1) You do this by finding the X and Y components of each using

the sine and cosine of the west and 60º south directions.

X_x = |X|cos(ø) and X_y = |X|sin(ø)

Don't confuse the X for position with the _x and _y subscripts for the horizontal and vertical !

2) Add up the X horizontal and Y vertical components

3) Then find the magnitude of the resultant with Pythagoras' Theorem

-- Don't worry about the angle since you are only asked for the magnitude.

Then you can use the definition of velocity --- v(avg) = ∆x / ∆t --- to find the velocity.