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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