Find an expression for v in terms of t .

Sketch the graph of v against t , clearly showing the coordinates of any intercepts, and the equations of any asymptotes.

(i)     Write down the time T at which the velocity is zero.

(ii)     Find the distance travelled in the interval [0, T] .

Find an expression for s , the displacement, in terms of t , given that s = 0 when t = 0 .

Hence, or otherwise, show that $s = \frac{1}{2}\ln \frac{2}{{1 + {v^2}}}$.

$\frac{{{\text{d}}v}}{{{\text{d}}t}} = - {v^2} - 1$

attempt to separate the variables     M1

$\int {\frac{1}{{1 + {v^2}}}{\text{d}}v = \int { - 1{\text{d}}t} }$     A1

$\arctan v = - t + k$     A1A1 

Note: Do not penalize the lack of constant at this stage.

 

when t = 0, v = 1     M1

$\Rightarrow k = \arctan 1 = \left( {\frac{\pi }{4}} \right) = (45^\circ )$     A1

$\Rightarrow v = \tan \left( {\frac{\pi }{4} - t} \right)$     A1

[7 marks]

     A1A1A1

 

Note: Award A1 for general shape,

   A1 for asymptote,

   A1 for correct t and v intercept.

 

Note: Do not penalise if a larger domain is used.

 

[3 marks] 

(i)     $T = \frac{\pi }{4}$     A1

(ii)     area under curve $= \int_0^{\frac{\pi }{4}} {\tan \left( {\frac{\pi }{4} - t} \right){\text{d}}t}$     (M1)

$= 0.347\left( { = \frac{1}{2}\ln 2} \right)$     A1

[3 marks] 

$v = \tan \left( {\frac{\pi }{4} - t} \right)$

$s = \int {\tan \left( {\frac{\pi }{4} - t} \right){\text{d}}t}$     M1

$\int {\frac{{\sin \left( {\frac{\pi }{4} - t} \right)}}{{\cos \left( {\frac{\pi }{4} - t} \right)}}} {\text{ d}}t$     (M1)

$= \ln \cos \left( {\frac{\pi }{4} - t} \right) + k$     A1

when $t = 0,{\text{ }}s = 0$

$k =  - \ln \cos \frac{\pi }{4}$     A1

$s = \ln \cos \left( {\frac{\pi }{4} - t} \right) - \ln \cos \frac{\pi }{4}\left( { = \ln \left[ {\sqrt 2 \cos \left( {\frac{\pi }{4} - t} \right)} \right]} \right)$     A1

[5 marks]

METHOD 1

$\frac{\pi }{4} - t = \arctan v$     M1

$t = \frac{\pi }{4} - \arctan v$

$s = \ln \left[ {\sqrt 2 \cos \left( {\frac{\pi }{4} - \frac{\pi }{4} + \arctan v} \right)} \right]$

$s = \ln \left[ {\sqrt 2 \cos (\arctan v)} \right]$     M1A1

 

 

$s = \ln \left[ {\sqrt 2 \cos \left( {\arccos \frac{1}{{\sqrt {1 + {v^2}} }}} \right)} \right]$     A1

$= \ln \frac{{\sqrt 2 }}{{\sqrt {1 + {v^2}} }}$

$= \frac{1}{2}\ln \frac{2}{{1 + {v^2}}}$     AG

 

METHOD 2

$s = \ln \cos \left( {\frac{\pi }{4} - t} \right) - \ln \cos \frac{\pi }{4}$

$= - \ln \sec \left( {\frac{\pi }{4} - t} \right) - \ln \cos \frac{\pi }{4}$     M1

$= - \ln \sqrt {1 + {{\tan }^2}\left( {\frac{\pi }{4} - t} \right)}  - \ln \cos \frac{\pi }{4}$     M1

$= - \ln \sqrt {1 + {v^2}}  - \ln \cos \frac{\pi }{4}$     A1

$= \ln \frac{1}{{\sqrt {1 + {v^2}} }} + \ln \sqrt 2$     A1

$= \frac{1}{2}\ln \frac{2}{{1 + {v^2}}}$     AG

 

METHOD 3

$v\frac{{dv}}{{ds}} = - {v^2} - 1$     M1

$\int {\frac{v}{{{v^2} + 1}}dv = - \int {1ds} }$     M1

$\frac{1}{2}\ln ({v^2} + 1) = - s + k$     A1

when $s = 0\,,{\text{ }}t = 0 \Rightarrow v = 1$

$\Rightarrow k = \frac{1}{2}\ln 2$     A1

$\Rightarrow s = \frac{1}{2}\ln \frac{2}{{1 + {v^2}}}$     AG

 

[4 marks]

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.