Question

In reaching her destination, a backpacker walks with an average velocity of $1m/s$, due west. This average velocity results, because she hikes for $6km$ with an average velocity of $3m/s$ due west, turns around, and hikes with an average velocity of $0.3m/s$ due east.How far to east did she walk (in kilometers)?

• A. $1.714$
  
• B. $2$
  
• C. $6$
  
• D. $3$
  

Answer 1

Answer: (A)

$1.714$

We define average velocity as the ratio of total travelled distance and time

The backpacker walks $6㎞=6000m$ due to west

The average velocity was $3㎧$

Then the traveled time for this motion is $t_1=\frac{6000}{3}=2000sec$

The she travels east, lets assume her traveled time due to east is $t_2$

Then the traveled distance due $t_0$ east is,

$s_2=0.3t_2$

Then the total traveled distance is $6000+s_2$ and the total traveled time is $t_1+t_2=2000+t_2$

Then the final average velocity is

$v_{av}=\frac{6000+s_2}{2000+t_2}=\frac{6000+0.3t_2}{2000+t_2}=1$

$\therefore t_2=5714sec$

$\therefore s_2=0.3t_2=0.3\times 5.714$

$=1714m$

$=1.714㎞$