a.

Distance of the rocket from origin=150

b.

c.

2.

a.

but

So,

Substituting and obtained in equations 1 and 2 into AB we obtain:

Optain all piont in this equations.

b.

Force

Torque by force about point A solve equation.

c.

Torque of w about point A is

W= mg j

From equation 2,

Then, from equation 3

d.

The resultant torque about point equation

The equation has two z components. That is, and

For the resultant torque to be negative, the latter component should be larger than the first component. That is,

f(2r-h)cos 0> f (sin0+ mg) 2 hr – h sqaure and root.

e.

We have:

Substituting the given variables into the inequality we get:

Dividing both sides by we obtain

The heaviest mass that can be pulled 73.215 kg

3.

2x +2y +4z =19000

2x +y +5z= 21000

2x+3y+4z=20000

We solve the equations using Gaussian elimination as follows:

(2x +y +5z) -(2x +2y +4z) =21000-19000

-y+z=2000

Subtracting equation 1 from equation 2 we obtain

-2y+z=2000

Then, subtracting equation 3 from equation 2 we get

2000-1000= 1000

y =1000

Subtracting equation 5 from equation 4 we obtain

Substituting into equation 5 we get

Substituting y and y into equation 1 we get

Therefore

b(i)

b(ii)

batch 1 total cost 19000

batch size 10 per cost batch 19250/5 =18250

batch size 10 per cost batch 183000/10=18300

The most cost effective batch size is 5 since it has the lowest unit cost.

4.

a.

When the position of the hypocycloid will be:

r(t) has an x-component only implying that the hypocycloid is a horizontal segment.

b.

Given that

c.

when r =r2

acce r”(t)

Hence, the acceleration is proportion to the distance oP