vector problem- trouble with angles

• Sep 2nd 2009, 03:29 PM
BiGpO6790
vector problem- trouble with angles
A car is driven east for a distance of 45 km, then north for 29 km, and then in a direction 29° east of north for 27 km. Determine (a) the magnitude (in km) of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.

I got the resultant but i cannot figure out which angle i am trying solve for. please help
• Sep 2nd 2009, 07:09 PM
Soroban
Hello, BiGpO6790!

Quote:

A car is driven east for a distance of 45 km, then north for 29 km,
and then in a direction 29° east of north for 27 km.

Determine:

(a) the magnitude (in km) of the car's total displacement from its starting point

(b) the angle (from east) of the car's total displacement measured from its starting direction.

Code:

                    N    * D                     :    /:                     :29°/ :                     :  /27: y                     : /  :                     :/61° :                   C * - - * E                     |  x  :                     |    :                 29 |    : 29                     |    :     A * - - - - - - * - - *             45      B  x  F

The car started at $\displaystyle A$, drove east 45 km to $\displaystyle B$,
then north 29 km to $\displaystyle C$, then at heading N29°E for 27 km to $\displaystyle D.$

We have: .$\displaystyle AB = 45,\:BC = EF = 29,\:CD = 27$
. . $\displaystyle \angle NCD = 29^o \quad\Rightarrow\quad \angle DCE = 61^o$

Let $\displaystyle x = CE = BF,\:y = DE$

Draw line segment $\displaystyle AD.$
Its length is the displacement in part (a).
. . In right triangle $\displaystyle DFA\!:\;\;AD \:=\:\sqrt{(x+45)^2 + (y+29)^2}$ .[1]

In right triangle $\displaystyle DEC\!:\;\;\begin{Bmatrix}x \:=\:27\cos61^o \:\approx\:13.09 \\ y \:=\:27\sin61^o \:\approx\:23.61\end{Bmatrix}$

Substitute into [1]: /$\displaystyle AD \:=\:\sqrt{(13.09+ 45)^2 + (23.61 + 29)^2} \:=\:\sqrt{58.09^2 + 52.61^2} \:\approx\:78.37\text{ km}$ .(a)

We want: .$\displaystyle \theta = \angle DAF$

$\displaystyle \tan\theta \:=\:\frac{DF}{AF} \:=\:\frac{52.61}{58.09} \:=\:0.905663625$

Therefore: .$\displaystyle \theta \;\approx\;42.17^o$ .(b)