Please Help Me with Vectors Application---I need Answers--year 12 MATHS--please answer them -if you can --?

Question

Q1. Peter walks from her house for 2 km in a direction N15`E and then turns right through 90 degrees to the original directions.She than walks for a further distance of 1 km to arrive at Stephen house.
a) Find the magnitude od the resultant displacement
b)find the angle between this vector and Peter`s original directions
c) What is the direction of stephen house from Peters house?
Q2) A Emirates aeroplane takes off from Tullamarine airport bound for Cairns,and flies for 20 km at an angle of elevation of 30*, and horizontally at 200 km/h . Find the displacemtn from thr take off point after six minutes of horizontaly flying.
Q3)During the Gipspsland floods Ron rows his boat with a velocity od 6km/h directly towards the opposite bank of a river which is flowing at 2km/h A)Find the velocityof the boat relative to the bank of the river. and if the river is 100m wide how far down down the river below the will Rons boat reach the oppositebank

Answer 1

Q1. Draw it out. You have described a right triangle.
a. Magnitude = SQRT(2^2 + 1^2) = SQRT(5)

b. tan(Angle) = 1/2
Angle = 26.57 degrees

Q2. There are two ways of looking at this. The first is that the plane travels a horizontal distance of 20 km while moving upwards at a 30 degree angle. The second is that the plane flies 20 km along the 30 degree slope. There are also different ways that the displacement can be looked at: either horizontal or along the diagonal from the take off point to the point reached. I will do both.

This is assuming the 20 km is horizontal.

The plane reaches an altitude of:
H = 20tan(30) = 20SQRT(3)/3

In 6 mins of horizontal flight, the plane goes a distance of:
D = (6/60)200 = 20 km

Add this 20 km to the o0riginal 20 km and get a total horizontal distance covered of 40 km.

This horizontal distance along with the altitude forms a right triangle so c^2 = a^2 + b^2 can be used.

Diagonal displacement = SQRT(40^2 + H^2)
= SQRT(1600 + 400/3) = 41.63 km

If we consider the other case, where the plane flies 20 kilometers along a 30 degree slope the:

Ground Distance = 20cos(30) = 10SQRT(3)
The plane flies horizontal for 20 km (as above) so:
total horizontal distance = 20 + 10SQRT(3) = 37.32 km

The diagonal displacement.

H = 20sin(30) = 10

Diagonal displacement = SQRT(10^2 + 37.32^2)
= 38.64 km

Q3. Use c^2 = a^2 + b^2 since the velocities form a right triangle.
Velocity = SQRT(6^2 + 2^2) = SQRT(40) = 6.33 km/h

The velocity perpendicular to the river is 6 km/h and the river is 100 m (or .1 km) wide so the time taken:
Time = 0.1/6 hours = 1/60 hours = 1 minute

Distance down stream: (1/60)2 = 1/30 km = 33.333 meters

You can also do this just by ratios: D/100 = 2/6 so D = 200/6 = 33.333 (the ratio of the distance downstream to across will be the same as the velocoties in the corresponding direction since the times are the same - d1 = 2T and d2 = 6T and D/100 = d1/d2 = 2T/6T = 2/6)

Answer 2

Q1.
Vector one 2km at cartesian heading of 75 degrees.
Vector two 1km at cartesian heading of -15 degrees (90 degrees right of 15.)

Decompose:
x1 = 2cos(75). y1 = 2sin(75)
x2 = 1cos(-15) y2 = 1sin(-15)

a) magnitude by pythagorean (it had a 90 ange) =
sqrt( 1^2 + 2^2) = sqrt(5)
b) angle to Stephan's house from Peter's(origin) =
invTan(y/x) = invTan( (y1 + y2)/(x1 + x2))
But, the angle between this vector and his original heading is:
75 - invTan(y/x)
c) See the first part of b.

Q2
Vector 1 = 20 km at 30 degrees up
up = 20 sin(30degrees) = 15
horizontal = 20 cos(30) = 10sqrt(3)

vector 2 = 6 min/60 min/hr * 200 km/h horizontal
= 20 km horizontal

result = 15 km up and 20+10sqrt(3) horiz.

Q3 His velocity relative to the bank = sqrt(6^2 + 2^2)
= sqrt(40) = 2sqrt(10)
at an angle of invTan(6/2)

His distance downstream will be 100/6 *2 = 33.33 m