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

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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

1 Answer

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Answer 1 · 2017-02-02T14:11:13

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

Explanation:

(1) As hiker moves $6.0$ $6.0$ km. east and then $8.0$ $8.0$ km. north, his movement can be depicted in following figure.
enter image source here
Now as p Pythagoras theorem, his movement is $\sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$ $sqrt(6^2+8^2)=sqrt(36+64)=sqrt100=10$ km.

(2) Again the movement of plane is shown in following diagram. The plane travels north at $250$ $250$ $\frac{k m .}{h r .}$ $(km.)/(hr.)$ for two hour and is then $500$ $500$ $k m .$ $km.$ north, but is displaced $40$ $40$ $k m .$ $km.$ to southwest.
enter image source here
The resultant displacement is given by solution of triangle for third side i.e. $\sqrt{500^{2} + 40^{2} - 2 \times 40 \times 500 \times {cos 135}^{\circ}}$ $sqrt(500^2+40^2-2xx40xx500xxcos135^@)$

= $\sqrt{250000 + 1600 + 40000 \times 0.7071}$ $sqrt(250000+1600+40000xx0.7071)$

= $\sqrt{279884} = 529.04$ $sqrt279884=529.04$

and resultant velocity is $\frac{529.04}{2} = 264.52$ $529.04/2=264.52$ $\frac{k m}{h r .}$ $(km)/(hr.)$ towards east at an angle $θ$ $theta$ from north given by $\frac{sin θ}{40} = \frac{{sin 135}^{\circ}}{529.04}$ $sintheta/40=sin135^@/529.04$

i.e. $sin θ = \frac{40 \times 0.7071}{529.04}$ $sintheta=(40xx0.7071)/529.04$ and $θ = {3.065}^{\circ}$ $theta=3.065^@$

(3) As Andy can paddle canoe at $1.2$ $1.2$ $\frac{m}{s}$ $m/s$ , he covers $200 m$ $200m$ in $\frac{200}{1.2}$ $200/1.2$ seconds. During this period river current sends him $\frac{200}{1.2} \times 0, 4 = \frac{200}{3} m$ $200/1.2xx0,4=200/3m$ east.
enter image source here
As such he should paddle towards west and as distance covered is

$\sqrt{200^{2} + {(\frac{200}{3})}^{2}} = \frac{200}{3} \sqrt{10} = 210.82$ $sqrt(200^2+(200/3)^2)=200/3sqrt10=210.82$

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

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

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Impact of this question

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