IB DP Math AI: Topic SL 5.13: Kinematic problems: IB style Questions HL Paper 2

Question 7. [Maximum mark: 17]

Consider the following system of coupled differential equations.

\(\frac{dx}{dt}\)= −4x

\(\frac{dx}{dt}\)= 3x − 2y

a. Find the eigenvalues and corresponding eigenvectors of the matrix  [6]

b.  Hence, write down the general solution of the system. [2]

c.  Determine, with justification, whether the equilibrium point (0, 0) is stable or unstable. [2]

d.  Find the value of \(\frac{dy}{dx}\)

(i) at (4 , 0).

(ii) at (−4, 0). [3]

e.  Sketch a phase portrait for the general solution to the system of coupled differential equations for −6 ≤ x ≤ 6, −6 ≤ y ≤ 6. [4]

▶️Answer/Explanation

(a)

\( (-4-\lambda )(-2-\lambda )=0\)

\(\lambda =-4\) OR \(\lambda =-2 \)

\(\lambda =-4\)

\(3x-2y=-4y\)

\(3x=-2y\) 

possible eigenvector is \((_{3}^{-2})(or any real multiple) \lambda =-2\) \(x = 0, y =  1\) possible eigenvector is \((_{1}^{0}) (or any real multiple)

(c) two (distinct) real negative eigenvalues (or equivalent (eg both \(e^{-4t}\rightarrow 0,e^{-2t}\rightarrow 0 as t\rightarrow \)∞ ⇒ stable equilibrium point

(d)\(\frac{dy}{dx}=\frac{3x-2y}{-4x}(i)(4,0)\Rightarrow \frac{dy}{dx}=-\frac{3}{4}(ii)(-4,0)\Rightarrow \frac{dy}{dx}=-\frac{3}{4}\)

(e)

Question 6. [Maximum mark: 15]

A particle P moves along the x-axis. The velocity of P is v m s1 at time t seconds, where

v = −2t2 + 16t − 24 for t ≥ 0.

a. Find the times when P is at instantaneous rest. [2]

b.  Find the magnitude of the particle’s acceleration at 6 seconds. [4]

c.  Find the greatest speed of P in the interval 0 ≤ t ≤ 6. [2]

d.  The particle starts from the origin O. Find an expression for the displacement of P from O at time t seconds. [4]

e.  Find the total distance travelled by P in the interval 0 ≤ t ≤ 4. . [3]

▶️Answer/Explanation

(a) solving v = 0 t= 2, t= 6

(b)use of power rule \(\frac{dv}{dt}=-4t+16 (t=6)\Rightarrow a=-8magnitude = 8 ms^{-2}\)

(c) using a sketch graph of \(v=24ms^{-1}\)

(d) \(x=\int vdt\) attempt at integration of

\(v -\frac{2t^{3}}{3}+8t^{2}-24t+c\)

attempt to find c (use of t x = 0 = x= 0)

\(c = 0\)      \(x=-\frac{2t^{3}}{3}+8t^{2}-24t)\)

(e) \(\int_{0}^{4}|v|dt = 32 m\) 

Question 6. [Maximum mark: 18]

An ice-skater is skating such that her position vector when viewed from above at time t

seconds can be modelled by

\((_{x}^{y})\) =

with respect to a rectangular coordinate system from a point O, where the non-zero constants a and b can be determined. All distances are in metres.

a.  Find the velocity vector at time t . [3]

b.  Show that the magnitude of the velocity of the ice-skater at time t is given by

                 \(a^{ebt}\sqrt{(1+b^{2})}\) [4]

At time t = 0 , the displacement of the ice-skater is given by \((_{0}^{5})\) and the velocity of the

ice-skater is given by ( 5 ) .

c.    Find the value of a and the value of b . [3]

d.  Find the magnitude of the velocity of the ice-skater when t = 2 . [2] At a point P, the ice-skater is skating parallel to the y-axis for the first time.

e.   Find OP. [6]

▶️Answer/Explanation

(a) use of product rule \((_{y}^{x})= (_{abe^{bt}sin t + ae^{bt}cost t}^{abe ^{bt}cost – ae ^{bt}sin t})\)

(b) \(|v|^{2}= x^{2}+y^{2}=[abe^{bt}cost – ae^{bt}sint]^{2}+ [abe^{bt}sint+ae^{bt}cost]^{2}\)

\(= [a^{2}sin^{2}t – 2a^{2}bsin t cost + a^{2}b^{2}cos^{2}t + a^{2}cos^{2}+ 2a^{2}bsint costt +a^{2}b^{2}sin^{2}t ]\) use of \(sin^{2}t + cos^{2}t=1\) 

 \(=a^{2}(b^{2}+1)e^{2bt}\) magnitude of velocity is  \(ae^{bt}\sqrt{(1+ b^{2})}\)

(c) when t = 0, \(ae^{bt}cost = 5 a=5 abe^{bt}cost – ae^{bt}sint = -3.5 b = -0.7\ )

(d)\(5e^{-0.7\times 2}\sqrt{(1+(-0.7)^{2})}1.51(1.50504…)\)

(e) \(x=0 ae^{bt}(b cost – simt)=0 tant=b t = 2.53 (2.53086…)\) correct substitution of their t to find x or y

\(x = − 0.697 (- 0.696591…)\) and \(y = 0.488 (0.487614…)\)

use of Pythagoras / distance formula \(OP = 0.850 m  (0.850297…)\)

Question

A particle moves in a straight line. The velocity, v ms1 , of the particle at time t seconds is given by  v (t) = t sin t – 3, for 0 ≤ t ≤ 10.

The following diagram shows the graph of v .

                 

    1. Find the smallest value of t for which the particle is at rest. [2]

    2. Find the total distance travelled by the particle. [2]

    3. Find the acceleration of the particle when t = 7 . [2]

▶️Answer/Explanation

Ans:

(a) recognising  v= o  t = 6.74416… = 6.74

= 37.0968…

= 37.1 (m)

(c)

recognizing acceleration at  t = 7  is given by v′(7)

acceleration = 5.93430… 

5.93

Question

Two boats A and B travel due north.

Initially, boat B is positioned 50 metres due east of boat A.

The distances travelled by boat A and boat B, after t seconds, are x metres and y metres respectively.

The angle θ is the radian measure of the bearing of boat B from boat A.

This information is shown on the following diagram.

    1. Show that y = x + 50 cot θ.  [1]

      At time T , the following conditions are true.

      Boat B has travelled 10 metres further than boat A. Boat B is travelling at double the speed of boat A.

      The rate of change of the angle θ is 0.1 radians per second.

    2. Find the speed of boat A at time T . [6]

▶️Answer/Explanation

Ans: 

(a)

tan \(\Theta = \frac{50}{y-x}\)

OR

\(cot \Theta = \frac{y-x}{50 }\)

\(y= x+50 \: cot\Theta \)

(b)

attempt to differentiate with respect to t

\(\frac{dy}{dt}= \frac{dx}{dt}= – 50 (cosec\Theta )^{2}\frac{d\Theta }{dt}\)

attempt to set speed of B equal to double the speed of A

\(2 \frac{dx}{dt}=\frac{dx}{dt} – 50 (cosec\Theta )^{2}\frac{d\Theta }{dt}\)

\(\Theta = arctan 5 (=1.373…= 78.69..)\)

OR

\(cosec ^{2}\Theta = 1+ cot^{2}= 1+(\frac{1}{5})^2= \frac{26}{25}\)

\(\frac{dx}{dt}= -50(\frac{26}{25})\times -0.1\)

So the speed of boat A is 5.2 (ms-1)

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