# A car drives north at 50 mi/h for 10 min and then turns east and goes 5.0 min at 62 mi/h. Finally, it goes southwest at 26 mi/h for 6.0 min. what is the car's displacement? What is its average velocity for this trip?

##### 1 Answer
Sep 28, 2016

$\text{please have a look the animation}$
${v}_{a} = 22.06 \text{ } m p h$

#### Explanation:

$\overline{A D} = 50 \cdot \frac{10}{60} = \frac{50}{6} = 8.33 \text{ miles}$

$\overline{D E} = 62 \cdot \frac{5}{60} = \frac{31}{6} = 5.17 \text{ miles}$

$\overline{E B} = 26 \cdot \frac{6}{60} = \frac{156}{60} = 2.6 \text{ miles}$

$\overline{A D} = \overline{E C}$
$\overline{D E} = \overline{A C}$

$\text{use ABC triangle}$

$\overline{A C} = \overline{D E} = 5.17$

$\overline{B C} = \overline{E C} - \overline{E B}$

$\overline{B C} = 8.33 - 2.6 = 5.73$

$\text{displacement} = \overline{A B} = \sqrt{{\overline{A C}}^{2} + {\overline{B C}}^{2}}$

$\overline{A B} = \sqrt{{\left(5.17\right)}^{2} + {\left(5.73\right)}^{2}}$

$\overline{A B} = 7.72 \text{ miles}$

"average velocity is described as " v_a=("displacement")/("time elapsed")

$\text{time elapsed="10/60+5/60+6/60=21/60=0.35 " } h$

${v}_{a} = \frac{\overline{A B}}{0.35}$

${v}_{a} = \frac{7.72}{0.35}$

${v}_{a} = 22.06 \text{ } m p h$