# Apparently simple canoe problem - or not?

## Homework Statement

A canoeist heads north for 1.2km on a lake and then heads west for 1.8km. Her speed is steady at 5m/s.
a) How long does it take her?
b) What is her average velocity?

## The Attempt at a Solution

1200m/(5m/s)=240s
1800m/(5m/s)=360s

Total of 600s but the answers suggested 0.60h

The average velocity was suggested as 3.6km/h or 1m/s but throughout her trip she held a constant speed of 5m/s.

So there is concern for the suggested answer for both a) and b)

Pivoxa, what book are these coming from? Are we outsourcing textbook writing these days as well? The first answer I agree with you.
If you assume they meant 5km/hr, .6hr fits. See if the second answer makes sense using this figure and calculating net displacement.

Pivoxa, what book are these coming from? Are we outsourcing textbook writing these days as well? The first answer I agree with you.
If you assume they meant 5km/hr, .6hr fits. See if the second answer makes sense using this figure and calculating net displacement.

They are looking for the average speed in b) so if we assume an average velocity of 5km/hr (which what they used in a)) then b) should be 5km/hr not 3.6km/hr.

I am using an Austrailan textbook from Nelson publishing company. There are many, many problems in it and errors may crop up at times.

They are looking for the average speed in b) so if we assume an average velocity of 5km/hr (which what they used in a)) then b) should be 5km/hr not 3.6km/hr.

I am using an Austrailan textbook from Nelson publishing company. There are many, many problems in it and errors may crop up at times.
well, needs a good dose of editing.............

Remember the definition of speed and velocity are very different in physics than in everyday language. For instance if you drive around a racetrack at an ave speed of 250Km/hr, your velocity between the green and checkered flags is still zero, 'cuz you haven't gone anywhwere, ie net displacement is zero. So in problem above, same applies, calculate actual displacement from starting point then divide by time to get velocity, which is a vector quantity (has direction associated with it). Speed is a scalar quantity, no direction implied.

First thing to do is draw a picture and think about the info given in terms of vectors and scalars...

The resultant of the two directions is 2163m - this is what you have to equate your average speed to - ie. 2163/5=0.6h

Last edited:
HallsofIvy
Homework Helper
No one can canoe at 5 m/s! I feel sure 5 km/h was intended. Then the total time was 0.60 hours. Remember that velocity, unlike speed, is a vector quantity. The magnitude of the velocity will be the straight line distance from starting point to ending point (use the Pythagorean theorem). You can use arctangent to get the angle.

D H
Staff Emeritus
No one can canoe at 5 m/s!

A very few select people can, for short distances. World-class times for the C-1 200 meter sprint are below 40 seconds (5 m/s), C-1 500 meter sprint, about 1:46 (4.7 m/s), K-1 500 meter sprint, about 1:38 (5.1 m/s).

No one can paddle at 5 m/s for any sustained period of time. Certainly not over a distance of 3 km.

I see. velocity is displacement/time

I see. velocity is displacement/time

And understanding this is a key to understanding much of physics--vectors have direction. scalars don't. Forces are vectors. Consider a merry go round, you can apply force in many directions and not get it to budge. Thats why these distinctions matter.

And understanding this is a key to understanding much of physics--vectors have direction. scalars don't. Forces are vectors. Consider a merry go round, you can apply force in many directions and not get it to budge. Thats why these distinctions matter.

What are you trying to illustrate with your example? Could you be clearer? That forces may balance and cancel hence get 0 displacement and 0 velocity? Speed would be 0 as well though.

Sorry, just trying to illustrate the importance of direction and why forces must have directions attached. You could push up, push down, push inwards toward the center, and the merry-go-round might deform, but not move. Apply a tangentially directed force to its axis of rotation and Voila!