Question

As she picks up her riders, a bus driver traverses four successive displacements represented by the expression

(-6.30b)i-(4.00bcos40⁰)i-(4.00bsin40⁰)j

+(3.00bcos50⁰)i-(3.00bsin50⁰)j-(5.00b)j

Here b represents one city block , a convenient unit of distance of uniform size; i is east and j is north . The displacements at40⁰and50⁰represents travel on road ways in the city that are at these angles to the main east-west and north-south streets.

a)What total distance did she travel ?
b)Compute the magnitude and direction of her total displacement.

Short answer

Total distance of travel by her = 18.3b

Magnitude of total displacement =12.4b

Direction of her total displacement =52.9⁰

Step-by-step solution

Definition of vector

A vector is a quantity that has both a direction and a magnitude, and is used to determine the relative location of two points in space.

Step 2:Formula

f vector A=a₁i+a₂j+a₃k and vector B=b₁i=b₂j=b₃k then the displacement of AB=(b₁-a₁)i+(b₂-a₂)j+(b₃-a₃)k

If vector A=x₁i=y₁j+z₁k and vector then B=x₂i =y₂j+z₂k the displacement of

The Pythagorean theorem and the definition of tangent will use for calculation

d=$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Distance Travel By Her

We will solve the total distance that she travelled since we have magnitude of 6.30b, 4b, 3b, 5b

So, total distance of d= 6.30b=4b+3b+5b= 18.3b

Magnitude Of Total Displacement

r= (-6.30b)i-(4.00bcos40⁰)i-(4.00bsin40⁰)j

+(3.00bcos50⁰)i-(3.00bsin50⁰)j-(5.00b)j

r= (-6.3b-4b$\times$0.77+3b$\times$0.64)i+(-4b$\times$0.64-3b$\times$0.77-5b)

r=_-7.5bi-9.9bj

$r=\sqrt{{(-7.5b)}^2+{(-9.9b)}^2}=b\sqrt{154.26}=12.4b$

Direction Of Total Displacement

The direction of her displacement

$$\begin{gathered} \tan \theta =\frac{9.9}{7.5} \\ \theta ={\tan }^{-1}1.32=52.9^0 \end{gathered}$$