If you could please help me with these problems I would really appreciate it very much. Thank you!

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e-devam | High School Teacher | (Level 2) Adjunct Educator

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15.

The resultant velocity of the boat is the vector sum of the velocities of the boat and that of the river, disposed at right angles to each other.

The river current was moving with a velocity of 3 ft/s, due South and the boat was moving with a velocity of 4 ft/s, due East.

The magnitude of the resultant can be found as follows:

`R=sqrt((4.0 ft.s^(-1))^2+(3.0 ft.s^(-1))^2)=5.0 ft.s^(-1)`

The direction of the resultant is the clockwise angle of rotation that the resultant vector makes with due East. This angle can be determined as follows (refer to the image):

`tantheta=3/4`

`theta=ATAN(3/4)`

= 36.9 degrees

So, the resultant velocity of the boat will be 5 ft/s at 36.9 degrees South of due East.

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thilina-g | College Teacher | (Level 1) Educator

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15.

The velocity of the river is 3 ft/s to south and the velocity of the boat is 4 feet per second to east.

Therefore the resultant velocity must lie in between south and east.

The angle can be found using the two velocities.

If the angle that the resultant velocity makes with east (the direction of velocity of boat) is `beta` ,

`tan beta = (3/4) = 0.75`

`beta = tan^-1 (0.75) = 36.9 degrees.`

The magnitude of the resultant velocity, v can be found by,

`v = sqrt(3^2+4^2) = sqrt(9+16) = sqrt(25) = 5 (ft/s)`

Therefore the resultant velocity is 5 ft/s in the direction of 36.9 degrees to south from east.

16.

A vector with length 2 and in the direction of <2,1> can be found by,

Let's first normalize the given vector (direction).

`u = 2i+j`

`|u| = sqrt(2^2+1^2)` `=sqrt(5)`

Now the unit vector is in the direction of <2,1> is given by,

`baru= (2i+j)/sqrt(5) `

Therefore the vector with length of 2 can be found by the scalar multiplication,

`v = 2. (2i+j)/sqrt(5)`

`v = 4/sqrt(5)i+2/sqrt(5)j`

The vector v can be given by `lt4/sqrt(5),2/sqrt(5)gt`

17.

Cube roots of 8 can be found by solving the following equation,

`x^3 -8 = 0 or x^3-2^3 =0`

We can rewrite this as,

`(x-2)(x^2+2x+2^2) = 0`

`(x-2)(x^2+2x+4) = `

One solution is (x-2) = 0 which gives x=2. That is the obvious solution.

There are two more solutions. They can be found by,

`(x^2+2x+4) = 0`

Using the quadratic formula,

`x = (-2+-sqrt(2^2-4xx1xx4))/(2xx1)`

`x = (-2+-sqrt(4-16))/(2)`

`x = (-2+-2sqrt(1-4))/(2)`

`x=-1+-sqrt(-3)`

This has two complex roots.

There are, `x = -1+sqrt(3)i or x = -1-sqrt(3)i`

Therefore, the cubic roots of 8 are, `2, -1+sqrt(3)i, and -1-sqrt(3)i`

18.

`3x^2+15y^2 = 15`

`x^2+4y^2 = 4`

Let `x^2=u and y^2=v`

Then,

`3u+15v = 15`

`u+4v = 4`

We get, u =4-4v from the second equation. By substituting that in the previous equation,

`3(4-4v)+15v = 15`

`12-12v+15v=15`

`3v = 3`

`v=1`

This gives u as,

`u = 4-4v = 4-4xx1 = 0`

`u=0 and v =1`

Therefore, `x^2 =0 and y^2=1`

This gives, `x =0` and

`y =+-sqrt(1) = +-1`

The solutions are, x =0, y=1 or y=-1.

19.

Start time = 7 am

The duration is 5 hours.

Upstream speed is 8 mph and downstream speed is 12 mph.

If the distance to the turning point is x, then,

`x/8+x/12 = 5`

`24xx(x/8+x/12) = 5xx24`

`3x+2x=120`

`5x = 120`

`x = 24`

Therefore, we should turn back after 24 miles.

The time to get there is, `24/8 = 3` hours.

Therefore we should turn back at (7+3), which is 10 am.

The answers are 10 am and 24 miles.

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