Vector calculus — Question 196

Original QCE Vault practice · 10 marks

Q196 · Practice questionTechnology-freeComplex unfamiliar10 marks

QUESTION 196 (10 marks)

A camera rig has velocity

\mathbf v(t)=2t\mathbf i+3t^2\mathbf j

m/s on

0\le t\le1

, and initial position (1,-1) m.

Determine its position vector.

[2 marks]

Determine the Cartesian tangent equation at t=1.

[2 marks]

Determine the exact distance travelled.

[3 marks]

The particle has mass 2 kg. Determine the resultant impulse over

0\le t\le1

and its magnitude. Explain why the magnitude cannot be obtained by treating the distance in (c) as a velocity change.

[3 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

10 marks

ANSWER

(a)

(1+t^2,-1+t^3)

. (b)

y=3(x-2)/2

. (c)

(13\sqrt{13}-8)/27

m. (d) Impulse

4\mathbf i+6\mathbf j

N s; magnitude

2\sqrt{13}

N s.

Worked solution

(a) Integrating the components gives

(t^2+C_1,t^3+C_2)

. Initial position requires

C_1=1,C_2=-1

(b) The point at t=1 is (2,0) and velocity (2,3) gives slope 3/2. Hence

y-0=(3/2)(x-2)

t\sqrt{4+9t^2}

. Thus

d=\int_0^1t\sqrt{4+9t^2}\,dt=\left[\frac{(4+9t^2)^{3/2}}{27}\right]_0^1=\frac{13\sqrt{13}-8}{27}.

Distance is the integral of speed rather than the magnitude of displacement.

(d) Impulse is change in vector momentum:

\mathbf J=m[\mathbf v(1)-\mathbf v(0)]=2[(2,3)-(0,0)]=(4,6)

N s. Its magnitude is

\sqrt{4^2+6^2}=2\sqrt{13}

. Distance integrates speed and has units of length; it is not the endpoint difference of vector velocities.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Integrate both components.

[1 mark]

Part a: Apply the initial position.

[1 mark]

Part b: Find the point and tangent direction.

[1 mark]

Part b: Give the tangent equation.

[1 mark]

Part c: Form the speed integral.

[1 mark]

Part c: Integrate by substitution.

[1 mark]

Part c: Evaluate the exact distance.

[1 mark]

Part d: Use change in vector momentum.

[1 mark]

Part d: Find the vector and its magnitude.

[1 mark]

Part d: Distinguish travelled distance from velocity change.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

View the QCAA syllabus

How many marks did you earn?

Compare your working with the guide above.

Vector calculus — Question 196

QUESTION 196 (10 marks)

Practice marking scheme

Related questions