(i)     Find his acceleration $a(t)$ for $t < 10$.

(ii)     Calculate $v(10)$.

(iii)     Show that $s(10) = 500$.

At $t = 10$ his parachute opens and his acceleration $a(t)$ is subsequently given by $a(t) =  - 10 - 5v,{\text{ }}t \ge 10$.

Given that $\frac{{{\text{d}}t}}{{{\text{d}}v}} = \frac{1}{{\frac{{{\text{d}}v}}{{{\text{d}}t}}}}$, write down $\frac{{{\text{d}}t}}{{{\text{d}}v}}$ in terms of $v$.

You are told that Richard’s acceleration, $a(t) =  - 10 - 5v$, is always positive, for $t \ge 10$.

Hence show that $t = 10 + \frac{1}{5}\ln \left( {\frac{{98}}{{ - 2 - v}}} \right)$.

You are told that Richard’s acceleration, $a(t) =  - 10 - 5v$, is always positive, for $t \ge 10$.

Hence find an expression for the velocity, $v$, for $t \ge 10$.

You are told that Richard’s acceleration, $a(t) =  - 10 - 5v$, is always positive, for $t \ge 10$.

Find an expression for his height, $s$, above the ground for $t \ge 10$.

You are told that Richard’s acceleration, $a(t) =  - 10 - 5v$, is always positive, for $t \ge 10$.

Find the value of $t$ when Richard lands on the ground.

(i)     $a(t) = \frac{{{\text{d}}v}}{{{\text{d}}t}} =  - 10{\text{ (m}}{{\text{s}}^{ - 2}}{\text{)}}$     A1

(ii)     $t = 10 \Rightarrow v =  - 100{\text{ (m}}{{\text{s}}^{ - 1}}{\text{)}}$     A1

(iii)     $s = \int { - 10t{\text{d}}t =  - 5{t^2}( + c)}$     M1A1

$s = 1000{\text{ for }}t = 0 \Rightarrow c = 1000$     (M1)

$s =  - 5{t^2} + 1000$     A1

at $t = 10,{\text{ }}s = 500{\text{ (m)}}$     AG

 

Note:     Accept use of definite integrals.

[6 marks]

$\frac{{{\text{d}}t}}{{{\text{d}}v}} = \frac{1}{{( - 10 - 5v)}}$     A1

[1 mark]

METHOD 1

$t = \int {\frac{1}{{ - 10 - 5v}}{\text{d}}v =  - \frac{1}{5}\ln ( - 10 - 5v)( + c)}$     M1A1

 

Note:     Accept equivalent forms using modulus signs.

 

$t = 10,{\text{ }}v =  - 100$

$10 =  - \frac{1}{5}\ln (490) + c$     M1

$c = 10 + \frac{1}{5}\ln (490)$     A1

$t = 10 + \frac{1}{5}\ln 490 - \frac{1}{5}\ln ( - 10 - 5v)$     A1

 

Note:     Accept equivalent forms using modulus signs.

 

$t = 10 + \frac{1}{5}\ln \left( {\frac{{98}}{{ - 2 - v}}} \right)$     AG

 

Note:     Accept use of definite integrals.

 

METHOD 2

$t = \int {\frac{1}{{ - 10 - 5v}}{\text{d}}v =  - \frac{1}{5}\int {\frac{1}{{2 + v}}{\text{d}}v =  - \frac{1}{5}\ln \left| {2 + v} \right|( + c)} }$     M1A1

 

Note:     Accept equivalent forms.

 

$t = 10,{\text{ }}v =  - 100$

$10 =  - \frac{1}{5}\ln \left| { - 98} \right| + c$     M1

 

Note:     If $\ln ( - 98)$ is seen do not award further A marks.

 

$c = 10 + \frac{1}{5}\ln 98$     A1

$t = 10 + \frac{1}{5}\ln 98 - \frac{1}{5}\ln \left| {2 + v} \right|$     A1

 

Note:     Accept equivalent forms.

 

$t = 10 + \frac{1}{5}\ln \left( {\frac{{98}}{{ - 2 - v}}} \right)$     AG

 

Note:     Accept use of definite integrals.

[5 marks]

$5(t - 10) = \ln \frac{{98}}{{( - 2 - v)}}$

$\frac{{2 + v}}{{98}} =  - {{\text{e}}^{ - 5(t - 10)}}$     (M1)

$v =  - 2 - 98{{\text{e}}^{ - 5(t - 10)}}$     A1

[2 marks]

$\frac{{{\text{d}}s}}{{{\text{d}}t}} =  - 2 - 98{{\text{e}}^{ - 5(t - 10)}}$

$s =  - 2t + \frac{{98}}{5}{{\text{e}}^{ - 5(t - 10)}}( + k)$     M1A1

at $t = 10,{\text{ }}s = 500 \Rightarrow 500 =  - 20 + \frac{{98}}{5} + k \Rightarrow k = 500.4$     M1A1

$s =  - 2t + \frac{{98}}{5}{{\text{e}}^{ - 5(t - 10)}} + 500.4$     A1

 

Note:     Accept use of definite integrals.

[5 marks]

$t = 250{\text{ for }}s = 0$    (M1)A1

[2 marks]

Total [21 marks]

Parts (i) and (ii) were well answered by most candidates.

In (iii) the constant of integration was often forgotten. Most candidates calculated the displacement and then used different strategies, mostly incorrect, to remove the negative sign from $- 500$.

Surprisingly part (b) was not well done as the question stated the method. Many candidates simply wrote down $\frac{{{\text{d}}v}}{{{\text{d}}t}}$ while others seemed unaware that $\frac{{{\text{d}}v}}{{{\text{d}}t}}$ was the acceleration.

Part (c) was not always well done as it followed from (b) and at times there was very little to allow follow through. Once again some candidates started with what they were trying to prove. Among the candidates that attempted to integrate many did not consider the constant of integration properly.

In part (d) many candidates ignored the answer given in (c) and attempted to manipulate different expressions.

Part (e) was poorly answered: the constant of integration was often again forgotten and some inappropriate uses of Physics formulas assuming that the acceleration was constant were used. There was unclear thinking with the two sides of an equation being integrated with respect to different variables.

Although part (e) was often incorrect, some follow through marks were gained in part (f).