Question

A child walks 5,0 meters north, then 4,0 meters east, and finally 2,0 meters
south, What is the magnitude of the resultant displacement of the child after the
entire walk?
and also The speed of an object undergoing constant acceleration increases from
8,0 meters per second to 16,0 meters per second in 10, seconds, How far
does the object travel during the 10, seconds? i dont understand what the steps are

Answer 1

these are two seperate questions

Answer 2

Hi,
1)
if you draw a diagram of what is actually happening you will be able to understand what is happening much easier. Also displacement means distance from the starting point not the distance traveled if you didn't know already.

so


if you follow this diagram you can see how the child has walked, you can then make a triangle and work the displacement which in the case is the root of the sum of the two other sides. This is just Pythagoras a^2+b^2=c^2

2) The second question is a basic Suvat equation with:
s=Displacement (m)
u= Initial velocity (ms^-1)
v= Final velocity (ms^-1)
a= Acceleration (ms^-2)
t= time (s)

If you note down all the information from the question first the next step becomes more obvious so, writing down the information you are given:
u=8ms^-1
v=16ms^-1
t=10
s=?
a=?
s here is our target value and we can use the equation:
s=1/2(u+v)*t
another method you could use to arrive at the same answer which shows where this equation comes from is to use draw a velocity time graph as shown.


The area of that trapezium is the displacement of the object which you work out from 1/2(a+b) * h, with a and b being the sides (so 8ms^-1 and 16ms^-1) and the height being 10s, you have to realize it is a trapezium on it's side.

so your answer would be s= 1/2(8+16) * 10 = 120m

* = multiply by the way.

Hope that helped!!!