Question

A cyclist rides 4.0 km due west, then 12.0 km 33° west of north. From this point she rides 9.0 km due east. What is the final displacement from where the cyclist started (in km)? (Express your answer in vector form. Assume the +x-axis is to the east, and t

Answer 1

Answer:

$\mathbf{r = (-5.064 \ \hat i + 6.536 \ \hat j) km}$

Explanation:

Given that:

A cyclist rides 4.0 km due west, then 12.0 km 33° west of north. From this point she rides 9.0 km due east.

Let:

$r_1 = 4.0 \ km \ due \ west \\ \\ r_2 = 12.0 \ km \ \ \ \ \ \theta = 33^0 \ west \ of \ north \\ \\ r_3 = 9.0 \ km \ due \ east$

Assuming that:

east is the + x axis  and has a unit vector of $\hat i$

north is the +y axis and has a unit vector of $\hat j$

west is the - x axis and has a unit vector of $-\hat {i}$

south is the - y axis and has a unit vector of -$\hat j$

The displacement $r_x$ in a given direction of x can be expressed by the formula:

$r_x = \sum r_i$

$r_x = -4-12 cos (33)+9$

$r_x = - 5.064 \ km$

The displacement $r_y$ in a given direction of y can be expressed by the formula:

$r_y = \sum r_j$

$r_y = 12 \ \ * s in (33)$

$r_y = 6.536 \ km$

The final displacement can be expressed by the relation;

$r = r_x \hat i + r_y \hat j$

$\mathbf{r = (-5.064 \ \hat i + 6.536 \ \hat j) km}$