# Can anyone help confirm the answer I have to these 3 questions?

## 1) 10.0 km 2) Magnitude: 253km/h Direction: N(9)E 3) Northwest and 2.6 minutes

Feb 2, 2017

Please see below. I have somewhat different results, but you can follow the process and check.

#### Explanation:

(1) As hiker moves $6.0$ km. east and then $8.0$ km. north, his movement can be depicted in following figure.

Now as p Pythagoras theorem, his movement is $\sqrt{{6}^{2} + {8}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$ km.

(2) Again the movement of plane is shown in following diagram. The plane travels north at $250$ $\frac{k m .}{h r .}$ for two hour and is then $500$ $k m .$ north, but is displaced $40$ $k m .$ to southwest.

The resultant displacement is given by solution of triangle for third side i.e. $\sqrt{{500}^{2} + {40}^{2} - 2 \times 40 \times 500 \times \cos {135}^{\circ}}$

= $\sqrt{250000 + 1600 + 40000 \times 0.7071}$

= $\sqrt{279884} = 529.04$

and resultant velocity is $\frac{529.04}{2} = 264.52$ $\frac{k m}{h r .}$ towards east at an angle $\theta$ from north given by $\sin \frac{\theta}{40} = \sin {135}^{\circ} / 529.04$

i.e. $\sin \theta = \frac{40 \times 0.7071}{529.04}$ and $\theta = {3.065}^{\circ}$

(3) As Andy can paddle canoe at $1.2$ $\frac{m}{s}$, he covers $200 m$ in $\frac{200}{1.2}$ seconds. During this period river current sends him $\frac{200}{1.2} \times 0 , 4 = \frac{200}{3} m$ east.

As such he should paddle towards west and as distance covered is

$\sqrt{{200}^{2} + {\left(\frac{200}{3}\right)}^{2}} = \frac{200}{3} \sqrt{10} = 210.82$

and time taken is $\frac{210.82}{1.2} = 175.7$ second or $\frac{175.7}{60} = 2.93$ minutes.