Rates of change and differential equations — Question 14

QCAA 2022, Paper 2 · 5 marks

Q14 · 2022 · Technology-activeSimple familiar5 marks

QUESTION 14 (5 marks)

An object is moving in a straight line with an acceleration represented by the differential equation

\frac{dv}{dt}=-(4+v^2),

where

v

is the object’s velocity (m s

^{-1}

) over time

t

(s), where

t\ge0

, until it comes to rest.

Determine the general solution of the differential equation.

[3 marks]

The initial velocity of the object is

1.5\text{ m s}^{-1}

. Determine the time when the particle comes to rest.

[2 marks]

WORKED SOLUTION

5 marks

ANSWER

\frac12\tan^{-1}(v/2)=-t+c

; the particle comes to rest after approximately

0.32

Worked solution

a) Starting with

\frac{dv}{dt}=-(4+v^2),

separate variables:

\frac{1}{4+v^2}\,dv=-dt.

Integrating,

\frac12\tan^{-1}\left(\frac v2\right)=-t+c.

b) Given

v(0)=1.5

c=\frac12\tan^{-1}\left(\frac{1.5}{2}\right)\approx0.32.

When the particle comes to rest,

v=0

, so

\frac12\tan^{-1}(0)\approx-t+0.32,

which gives

t\approx0.32\text{ s}

correctly establishes a suitable integration result based on the separation of variables technique

[1 mark]

determines one side of the general solution in terms of

v

[1 mark]

determines the other side of the general solution in terms of

t

[1 mark]

determines an expression that represents the integration constant

[1 mark]

determines the time when the particle is at rest

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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