M1 May 04 Qs

A small boat S, drifting in da sea, is modelled as a particle moving in a straight line at constant speed. When first sighted at 900, S is at a point with position vectors (4i-6j) km relative to a fixed origin O. At 0945, S is at the point with position vector (7i-7.5j)km. At time t hours after 900, S is at the point with position vector s km.

a) calculate the bearing on which S is drifting.
b)Find an expression for s in terms of t

At 1000 a motor boat M leaves O and travels with constant velocity (pi+qj) km/h. Given that M interceps S at 1015,
c) calculate values p and q.

Reply 1

MRLX69

2

got this on paper, hold on....

Reply 2

MRLX69

2

a)

@=theta

Tan@=1.5/3
@=26.565
@+90 = 116.565

Bearing = 116 degrees

Reply 3

ziya_the_king

OP

MRLX69

a)

@=theta

Tan@=1.5/3
@=26.565
@+90 = 116.565

Bearing = 116 degrees

why do u add 90 degrees i just dont get it

Reply 4

Gaz031

15

A small boat S, drifting in da sea, is modelled as a particle moving in a straight line at constant speed. When first sighted at 900, S is at a point with position vectors (4i-6j) km relative to a fixed origin O. At 0945, S is at the point with position vector (7i-7.5j)km. At time t hours after 900, S is at the point with position vector s km.
a) calculate the bearing on which S is drifting.

Distance moved in 45mins = i(7-4) + j(-7.5--6) = i(3) + j(-1.5)
You can use some basic trigonometry here to calculate the angle the displacement makes with the north line (j axis).

b)Find an expression for s in terms of t

Distance moved in 45mins = 3i - 1.5j.
Distance moved in 15mins = i - (1.5/3) j
Distance moved in 60 mins = 4i - (4x1.5/3) j
Distance moved in 60 mins = 4i - 2j.
Distance moved in one hour (t hours) = 4i - 2j.
Initial displacement = 4i-6j.
Hence the displacement at time t hours:
4i - 6j + 4ti - 2tj [as distance moved in one hour varies depending on the number of hours]
s = i[4+4t] + j[-6-2t].

At 1000 a motor boat M leaves O and travels with constant velocity (pi+qj) km/h. Given that M interceps S at 1015,
c) calculate values p and q.

Find position of small boat at 1015 (1.25 hrs after 9:00). That must equal the position of the motor boat at 1015 (.25hrs after 10:00). I assume intercepts means meets. Hence the two positions are equal. You can equate the i and j components to find p and q repectively.

Edit: I don't really have a calculator to manipulate fractions for me right now so i'll just do it as best i can (without resorting to MS excel):
Position of small boat at 1015 is S when t=1.25:
s=i[4+4t] + j[-6-2t]
s=i[4+5] + j[-6-2.5]
s=i[9] + j[-8.5].
Position of motorboat .25 hrs after 10:00:
0.25(pi+qj) - there is no initial displacement as it starts from the origin.

Hence: 9i - 8.5j = 0.25pi + 0.25qj
9 = 0.25p
p = 9x4 = 36

-8.5 = 0.25q
q = -8.5x4 = -34j.

Reply 5

MRLX69

2

b)

Position vector = (original vector) + VelocityXtime

s = (4i-6j)+vt

v = d/t = [(7i-7.5j)-(4i-6j)]/ (3/4) = (4i-2j)Km h^-1

therefore: s= (4i-6j)+(4i-2j)t

Reply 6

Gaz031

15

ziya_the_king

why do u add 90 degrees i just dont get it