Question #8f515

2 Answers
Nov 15, 2016

Distance walked = #160m# Displacement = #70.7m#

Explanation:

You need to know the difference between distance and displacement.

Distance is probably easier to start with. It simply means "How far did you walk altogether?"

There were 3 different stages: 50m, then 80m, then 30m

#50+80+30= 160m# is the total distance walked.

Displacement means "How far away from the starting point was I when I stopped walking?#

You should realise that walking east cancels out some of the distance to the west.

#< ---------#
#" "80 m# west

#" "--- > <-----#
#" "30m# east #" "50 m# west

That means that your final position is the same as walking

50 m North and then 50 m West.

These lines form 2 sides of a right-angled isosceles triangle.

You need to know the length of the longest side - the slanted line which goes directly from start to finish. It is called the hypotenuse of the triangle.

To find its length, you can draw a scale diagram and measure the length of the longest side.

However, there is a method called "Pythagoras's Theorem" which you can use to calculate that length..

#"length"^2 =50^2 +50^2#

#"length"^2 =5000#

#"length" = sqrt5000#

#"length" = 70.7m#

This length is the displacement, because you end up 70.7 m away from the starting place.

Nov 15, 2016

Just a note about how to deal with #sqrt(5000)# without a calculator.

Explanation:

#sqrt(5000) ->sqrt(50xx100) -> sqrt(50xx10^2)#

But #sqrt(10^2)=10# so take #10^2# outside the root giving:

#10sqrt(50)#

............................................................................................................

Ok lets see if there are any squared numbers in 50 that we can also take outside the root.

#50=5xx10#

But #10 =2xx5# so we have

#50=5xx5xx2 = 5^2xx2#
...................................................................................................................

#10sqrt(50) = 10sqrt(5^2xx2) = 10xx5sqrt(2) = 50sqrt(2)#

We can not take this any further as 2 is a prime number

So #sqrt(5000) =50sqrt(2)" "# as an exact value.

As an approximate value this is 70.71 to 2 decimal places.