A person walks 27m east and then walks 31m at an angle 38° north of east. What is the magnitude of the total displacement. Answer in units of m

User Icon for Anonymous Anonymous answered
9 years ago

19.08

User Icon for Explain Bot Explain Bot answered
11 months ago

To find the magnitude of the total displacement, we can use vector addition and the Pythagorean theorem.

First, let’s break down the displacements into their x and y components.

The displacement of 27m to the east would have an x-component of +27m and a y-component of 0m, since it is in the east direction and there is no north or south component.

Next, we need to calculate the x and y components of the second displacement at an angle of 38° north of east.

To find the x-component of this displacement, we use cosine: cos(38°) = adjacent/hypotenuse = x-component/31m.

So, x-component = cos(38°) * 31m = 0.790 * 31m ≈ 24.49m (rounded to 2 decimal places).

To find the y-component of this displacement, we use sine: sin(38°) = opposite/hypotenuse = y-component/31m.

So, y-component = sin(38°) * 31m = 0.624 * 31m ≈ 19.31m (rounded to 2 decimal places).

Now, we have the x and y components for both displacements. We can add them together to get the total displacement.

The x-component of the total displacement = 27m + 24.49m = 51.49m (rounded to 2 decimal places).

The y-component of the total displacement = 0m + 19.31m = 19.31m (rounded to 2 decimal places).

Now, using the Pythagorean theorem, the magnitude of the total displacement is given by the formula: magnitude = sqrt((x-component)^2 + (y-component)^2).

So, magnitude = sqrt((51.49m)^2 + (19.31m)^2) ≈ sqrt(2653.64m^2 + 373.52m^2) ≈ sqrt(3027.16m^2) ≈ 55.01m (rounded to 2 decimal places).

Therefore, the magnitude of the total displacement is approximately 55.01m.