A particle moves in a straight line in a positive direction from a fixed point O.

The velocity v m \({{\text{s}}^{ - 1}}\) , at time t seconds, where \(t \geqslant 0\) , satisfies the differential equation

\[\frac{{{\text{d}}v}}{{{\text{d}}t}} = \frac{{ - v(1 + {v^2})}}{{50}}.\]

The particle starts from O with an initial velocity of 10 m \({{\text{s}}^{ - 1}}\) .

(a)     (i)     Express as a definite integral, the time taken for the particle’s velocity to decrease from 10 m \({{\text{s}}^{ - 1}}\) to 5 m \({{\text{s}}^{ - 1}}\) .

(ii)     Hence calculate the time taken for the particle’s velocity to decrease from 10 m \({{\text{s}}^{ - 1}}\) to 5 m \({{\text{s}}^{ - 1}}\) .

(b)     (i)     Show that, when \(v > 0\) , the motion of this particle can also be described by the differential equation \(\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{ - (1 + {v^2})}}{{50}}\) where x metres is the displacement from O.

(ii)     Given that v =10 when x = 0 , solve the differential equation expressing x in terms of v.

(iii)     Hence show that \(v = \frac{{10 - \tan \frac{x}{{50}}}}{{1 + 10\tan \frac{x}{{50}}}}\).

(a)     (i)     EITHER

Attempting to separate the variables     (M1)

\(\frac{{{\text{d}}v}}{{ - v(1 + {v^2})}} = \frac{{{\text{d}}t}}{{50}}\)     (A1)

OR

Inverting to obtain \(\frac{{{\text{d}}t}}{{{\text{d}}v}}\)     (M1)

\(\frac{{{\text{d}}t}}{{{\text{d}}v}} = \frac{{ - 50}}{{v(1 + {v^2})}}\)     (A1)

THEN

\(t = - 50\int_{10}^5 {\frac{1}{{v(1 + {v^2})}}} {\text{d}}v\,\,\,\,\,\left( { = 50\int_5^{10} {\frac{1}{{v(1 + {v^2})}}{\text{d}}v} } \right)\)     A1     N3

 

(ii)     \(t = 0.732{\text{ (sec)}}\,\,\,\,\,\left( { = 25\ln \frac{{104}}{{101}}(\sec )} \right)\)     A2     N2

[5 marks]

 

(b)     (i)     \(\frac{{{\text{d}}v}}{{{\text{d}}t}} = v\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     (M1)

Must see division by v \((v > 0)\)     A1

\(\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{ - (1 + {v^2})}}{{50}}\)     AG     N0

 

(ii)     Either attempting to separate variables or inverting to obtain \(\frac{{{\text{d}}x}}{{{\text{d}}v}}\)     (M1)

\({\frac{{{\text{d}}v}}{{1 + {v^2}}} = - \frac{1}{{50}}\int {{\text{d}}x} }\) (or equivalent)     A1

Attempting to integrate both sides     M1

\(\arctan v = - \frac{x}{{50}} + C\)     A1A1

Note: Award A1 for a correct LHS and A1 for a correct RHS that must include C.

 

When \(x = 0{\text{ , }}v = 10{\text{ and so }}C = \arctan 10\)     M1

\(x = 50(\arctan 10 - \arctan v)\)     A1 N1

 

(iii)     Attempting to make \(\arctan v\) the subject.     M1

\(\arctan v = \arctan 10 - \frac{x}{{50}}\)     A1

\(v = \tan \left( {\arctan 10 - \frac{x}{{50}}} \right)\)     M1A1

Using tan(A − B) formula to obtain the desired form.     M1

\(v = \frac{{10 - \tan \frac{x}{{50}}}}{{1 + 10\tan \frac{x}{{50}}}}\)     AG     N0

[14 marks]

Total [19 marks]

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