QCE Specialist Maths: Vector calculus Q254

Q254 · Practice questionTechnology-activeVery complex unfamiliar12 marks

QUESTION 254 (12 marks)

A river is 100 m wide. Coordinates have

x

east and

y

north, with the near bank

y=0

and far bank

y=100

. A rescue boat starts at

(0,0)

and keeps a fixed velocity

(u_x,u_y)

relative to the water, with

u_x^2+u_y^2=25

and

u_y>0

. The current at northward coordinate

y

0.12y

m/s east. A raft moves east along the far bank from

(60,100)

0.5

m/s. Ignore acceleration while the boat sets its heading.

Cartesian river sketch with banks at y=0 and y=100, boat at the origin and raft at (60,100). Eastward arrows increase with y. The raft direction is indicated separately; no rescue heading is supplied.

Derive the boat path

x(y)

and its bank-crossing time in terms of

u_x,u_y

[3 marks]

Determine all fixed headings that intercept the raft on the far bank. Give both velocity pairs to three decimal places and show that you have found every admissible heading.

[4 marks]

Select the heading that gives the earliest interception. Give the time and the angle west of north to two decimal places.

[2 marks]

For this faster heading, determine the greatest distance the boat moves west of its starting line before turning east. Give the distance to two decimal places.

[3 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

12 marks

ANSWER

(a)

x(y)=\frac{u_x}{u_y}y+\frac{0.06}{u_y}y^2

;

T=100/u_y

. (b)

(u_x,u_y)=(-4.898,1.003)

(-3.190,3.850)

m/s. (c)

T=25.97

s; heading

39.64^{\circ}

west of north. (d)

11.01

m west.

Worked solution

(a) The ground velocity is

(u_x+0.12y,u_y)

. Hence

y=u_yt

and

x=u_xt+0.06u_yt^2

. Eliminate

t=y/u_y

to get the path and use

y=100

for the crossing time.

(b) At crossing,

x=100(u_x+6)/u_y

, whereas the raft is at

x=60+50/u_y

. Equality gives

u_x-0.6u_y=-5.5

. Combining this with the speed circle gives

136u_y^2-660u_y+525=0

. Thus

u_y=(165\pm25\sqrt{15})/68

, both positive, and

u_x=0.6u_y-5.5

. The two roots produce the stated pairs. The line and circle have exactly these two intersections, so there are no other admissible fixed headings.

u_y

increases. Select

u_y=3.8503615\ldots

u_x=-3.1897831\ldots

. Then

T=100/u_y=25.971587\ldots

and the west-of-north angle is

\tan^{-1}(-u_x/u_y)=39.639618\ldots{}^{\circ}

(d) The horizontal velocity changes sign at

y_*=-u_x/0.12=26.5815\ldots

m, which lies within the river. Substituting into the path gives

x_{\min}=-u_x^2/(0.24u_y)=-11.010564\ldots

m. Horizontal velocity is negative before and positive after this point, so this is the greatest westward excursion.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Form both ground-velocity components.

[1 mark]

Part a: Integrate and eliminate time to obtain the path.

[1 mark]

Part a: Obtain the crossing time.

[1 mark]

Part b: Use simultaneous boat and raft positions to obtain the heading line.

[1 mark]

Part b: Combine it with the speed constraint to obtain the quadratic.

[1 mark]

Part b: Calculate both velocity pairs.

[1 mark]

Part b: Check positive northward velocity and explain completeness.

[1 mark]

Part c: Select the larger northward component and calculate the earliest time.

[1 mark]

Part c: Calculate the correctly oriented heading angle.

[1 mark]

Part d: Find the turning coordinate from horizontal velocity.

[1 mark]

Part d: Evaluate the minimum horizontal coordinate.

[1 mark]

Part d: Check that it lies during the crossing and interpret the westward distance.

[1 mark]

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Vector calculus — Question 254

QUESTION 254 (12 marks)

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