Question

In reaching her destination, a backpacker walks with an average velocity of 1.34 m/s, due west. This average velocity results because she hikes for 6.44 km with an average velocity of 2.68 m/s, due west turns around, and hikes with an average velocity of 0.447 m /s, due east. How far east did she walk

Answer 1

We can define average velocity as:

av = (total distance travelled)/(total time)

Only using the above equation, we will find that she walked 807.3m due east.

The given information is:

The average velocity is 1.34 m/s due west.

(we can define west as the positive side and east as the negative side).

We know that first, she hikes 6.44km due west with an average velocity of 2.68 m/s.

To get the total time it took we write the equation:

2.68 m/s = (6,440m)/(time)

time = 6,440m/(2.68 m/s) = 2,402.9 seconds.

Then she hikes a distance D due east with an average velocity of 0.447 m/s, because she goes due east, we will write -D in the equations.

Now using the same equation as before, we can get the time as:

time' = -D/(-0.447 m/s) = D/(0.447 m/s)

Now the equation for the total average velocity will be:

1.34 m/s = (6,440m - D)/(2,402.9 s +  D/(0.447 m/s))

Now we need to solve this for D.

(1.34 m/s)*(2,402.9 s +  D/(0.447 m/s)) =  (6,440m - D)

(1.34 m/s)*(2,402.9 s) + (1.34 m/s)*(D/(0.447 m/s) =6,440m - D

(1.34 m/s)*(D/(0.447 m/s) + D = 6,440m - (1.34 m/s)*(2,402.9 s) = 3,229.1 m

D*(1 + (1.34m/s)/(0.447 m/s)) = 3,229.1 m

D*4 = 3,229.1 m

D =  3,229.1 m/4 = 807.3 m

We can conclude that she walked 807.3m due east.

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Answer 2

We have that the total distance traveled by the  backpacker is

$d_t=5.6km$

From the Question we are told that

Total Average velocity $V_{avg}= 1.34 m/s,$

Distance $d=6.44 km$

First average velocity $V_{1 avg}= 2.68 m/s,$

Second average velocity $V_{2 avg}= 0.447 m /s$

Generally the equation for Total Distance  is mathematically given as

$d_t=d_1+d_2\\\\V_tT_t=(v_1t_1)+(v_2t_2)$

Where

$t_1=\frac{d_1}{t_1}\\\\t_1=\frac{6440m}{2.68}\\\\t_1=2400s$

Therefore

$V_tT_t=(v_1t_1)+(v_2t_2)$

$1.34(T_t)=(2.68(2400))+(-(0.447t_2))$ ...(due to change in direction)

$1.34t_1+1.34t_2=6440-0.447t_2$

$t_2=1804s$

Hence, the distance covered by her is

$d_t=d_1+d_2\\\\d_t=(2.68(2400))+(-(0.447(1804)))\\\\d_t=5625.612m$

$d_t=5.6km$

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