m1 mechanics exam question help?

Hi Aliyah,

I'll go through everything and see what I can do for you (=

I'll rush through parts (a) and (b) as you know what you're doing, but I haven't watched the video so I may not be using the same notation etc.

You are given that:

r(9) = [-2 ; -4] = -2 [1 ; 2].

r(9⅔) = [4 ; -6] = 2 [2 ; -3].

r(9 + t) = [s{x} ; s{y}].

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(a)

For the bearing, use the difference in the two position vectors r(9) and r(9⅔):

r(9⅔) - r(9).

= 2 [2 ; -3] - -2 [1 ; 2].

= 2 [2 + 1 ; -3 + 2].

= 2 [3 ; -1].

This describes the movement vector over the ⅔ hour (or 40 minutes). To find the bearing, I will use:

180 - arctan(|r{x}| / |r{y}|).

= 180 - arctan(|6| / |-2|).

= 180 - arctan(6/2).

= 180 - arctan(3).

= 180 - 71.565051177...

= 108.434948822...

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(b)

To express s in terms of t, you need to recall that velocity is displacement over time.

You know the displacement over ⅔ hour is 2 [3 ; -1].

Dividing by ⅔ gives the displacement over 1 hour: (3/2) 2 [3 ; -1] = 3 [3 ; -1].

So, since distances are in km, and the time period is one hour, you have:

Velocity = 3 [3 ; -1] km h⁻¹.

Hence, the ship S moves this far per time unit t. Meaning, from displacement equals velocity multiplied by time, plus initial position:

s(t + 9) = r(9) + vt.

= -2 [1 ; 2] + 3 [3 ; -1] t.

= [9t - 2 ; -(3t + 4)].

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(c)

Extra information:

The ship M is introduced at 1100 hours at the origin O(0,0), or symbolically:

r{M}(11) = [0 ; 0]

The ship M has velocity pi + qj, or symbolically:

v{M} = [p ; q].

The ships M and S intercept, that is, have the same position, at 1130 hours, or time 11½, symbolically:

r{M}(11½) = s(11½).

To use the formula from before, we have displacement equal to the product of speed and time, plus initial position:

r{M}(t + 11) = r{M}(11) + v{M} t.

= [0 ; 0] + [p ; q] t.

= [p ; q] t.

= t [p ; q].

We know that s(t + 9) = [9t - 2 ; -(3t + 4)], so we can find s(11½) by substituting in t = 2½:

s(11½)

= [9*2½ - 2 ; -(3*2½ + 4)].

= [22½ - 2 ; -(7½ + 4)].

= [20½ ; -11½].

= ½ [41 ; -23].

Using the relation from the intersection, we have r{M}(11½) = s(11½), but we now know s(11½):

r{M}(11½) = ½ [41 ; -23]

We also have a formula for r{M}(t + 11) = t [p ; q]. At t = ½, this equals s(11½):

½ [p ; q] = ½ [41 ; -23].

Cancelling the ½ factor from both sides:

[p ; q] = [41 ; -23].

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I hope the math is clear enough. If not, feel free to message me. The key steps I have taken are:

[i] Note that s(2½ + 9) = r{M}(½ + 11) because of the interception.

[ii] Find a formula for r{M}(t + 11) using the information given and the formula r = r₀ + vt.

[iii] Evaluate s(11½) using s(t + 9) when t = 2½.

[iv] Equate s(11½) to r{M}(t + 11) when t = ½, using the formula from [ii].

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