Question

A person travel 20m along north then 10root2 along north east again cover 50m along west and then 12m along 30' south of west .FIND NET DISPLACEMENT.

Answer 1

plz help

Answer 2

Let positive x be east and positive y be north.

Net displacement it\[D_N = \sqrt{ x^2 + y^2}\]
Let's build up an expression for displacement in terms of our x (\(\hat i\)) and y (\(\hat j\)) displacements.

First, we travel 20m in positive y direction.
\(\therefore D = 20 \hat j\)

Second, we travel 100m at a 45 degree angle from the positive x-axis.
\(\therefore D = 20 \hat j + {\color{red} {100 \sin(45) \hat j + 100 \cos(45) \hat i}}\)
because we have a displacement in both the x and y direction.

Third, we travel 50 m in the negative x direction.
\(\therefore D = 20 \hat j + 100 \sin(45) \hat j + 100 \cos(45) \hat i - \color{red}{50 \hat i}\)

Finally, we travel 12 m at an angle of 30 degrees below the negative x-axis. We will have both a negative x and negative y displacement.
\(\therefore D = 20 \hat j + 100 \sin(45) \hat j + 100 \cos(45) \hat i - {50 \hat i}- \color{red}{12 \cos(30) \hat i - 12 \sin(30) \hat j}\)

Our final expression for displacement is
\[D = 20 \hat j + 10^2 \cos(45) \hat i + 10^2 \sin(45) \hat j + (-50) \hat i - 12 \cos(30) \hat i - 12 \sin(30) \hat j\]We can simply sum up the x and y components to find the total displacement in the x and y directions. We obtain \[x = 10^2 \cos(45) - 50 - 12 \cos(30)\]\[y = 20 + 10^2 \sin(45) - 12 \sin(30)\]

We can use these values in the first equation I presented.