Firstly, I would like to caution students regarding the use of solutions when preparing for exams. They are certainly a double-edged sword and since you are planning to invest an enormous amount of time, effort and money into getting a degree, it would be a good idea to use solutions properly from now on. In all cases you should only consult a solution once you think your answer is correct or if you have spent a significant amount of time trying to reach an answer. Coming up with your own conclusion after several minutes of independent thought is far more worthwhile than learning directly from the solution itself. You should also make sure you are critical of all solutions presented to you. Not only will this help your understanding, but solutions may have errors. Therefore if you find that you can't reconcile a solution with yours then take the time to get in touch with the author to discuss the discrepancy. For example, please do contact me if you have spotted a mistake below or would like further guidance on any of the questions.

I will firstly set out a series of hints. Try to do the questions from these alone and if you are still having difficulty then scroll down to the full solutions.

Hints

Section 2 - Physics

1 a) Work out the area enclosed by the function, taking any areas below the horizontal axis as negative.

1 b) This is a straight line therefore you need the gradient of it and any point on it.

1 c) (i) There is no need to integrate - consider what an integral calculates and use the graph.

1 c) (ii) This should be straightforward from your knowledge of velocity-time graphs.

1 d) Apply $F=ma$.

1 e) Apply $F=ma$ once more and then $P=Fv$

1 f) The graphs to be drawn are related to the one given by the area under it and the gradient of it.

2 a) Consider the force(s) acting on the stone and what their resultant can be equated to.

2 b) If the question seems unfamiliar to you then don't be put off and follow the instructions.

2 c) The initial velocity is $v_0$ and you have an expression for the velocity as a function of time. Let $v$ equal $v_0$ and work from there.

2 d) The most important aspect of both graphs is their behaviour for $t=0$ and as $t\rightarrow \infty$.

Full solutions Section 2 - Physics

1 a) Work out the area enclosed by the function, taking any areas below the horizontal axis as negative.

The areas enclosed by each triangle or rectangle, counting from left to right are as follows:

$3600$

$28800$

$4000$

$-4000$

$-28800$

$-3600$

If you add these together then you obtain $0$ so the train is back at the station.

1 b) This is a straight line therefore you need the gradient of it and any point on it.

The gradient is $\frac{-40}{200}=\frac{-1}{5}$ and a point is $(1100,0)$.

$v-0=\frac{-1}{5}(t-1100)$

$v=220-\frac{t}{5}$

1 c) (i) There is no need to integrate - consider what an integral calculates and use the graph.

The integral represents the area of the triangle $DE(1500,0)$ which is $-4000$.

1 c) (ii) This should be straightforward from your knowledge of velocity-time graphs.

This represents the distance travelled between $D$ and $E$.

1 d) Apply $F=ma$.

$a=\frac{-1}{5}$

$F=ma=2000N$

This force arises through braking and friction/resistance.

1 e) Apply $F=ma$ once more and then $P=Fv$

$F=ma=10000\times\frac{40}{180}=2222\frac{2}{9}$

$v=20$

$P=Fv=44444\frac{4}{9}W$

1 f) The graphs to be drawn are related to the one given by the area under it and the gradient of it.

2 a) Consider the force(s) acting on the stone and what their resultant can be equated to.

The only horizontal force on the stone is the drag force. It opposes the velocity; if $v$ is positive then the drag force should be negative so a minus sign is required in the equation. The resultant force is equal to $ma$ in accordance with Newton's second law.

2 b) If the question seems unfamiliar to you then don't be put off and follow the instructions.

2 b) (i) $[\frac{-1}{2v^2}]^v_{v_0}=\frac{1}{2v_0^2}-\frac{1}{2v^2}=\frac{-bt}{m}$

2 b) (ii) $\frac{1}{v^2}-\frac{1}{v_0}^2=\frac{2bt}{m}$

$\frac{1}{v^2}=\frac{2bt}{m}+\frac{1}{v_0^2}=\frac{2btv_0^2+m}{mv_0^2}$

$v^2=\frac{mv_0^2}{2btv_0^2+m}$

2 c) The initial velocity is $v_0$ and you have an expression for the velocity as a function of time. Let $v$ equal $\frac{v_0}{2}$ and work from there.

Let $v=\frac{v_0}{2}$

$\frac{v_0^2}{4}=\frac{mv_0^2}{2btv_0^2+m}$

$4m=2btv_0^2+m$

$t=\frac{3m}{2bv_0^2}$

2 d) The most important aspect of both graphs is their behaviour for $t=0$ and as $t\rightarrow \infty$.

$v=v_0\sqrt{\frac{m}{m+2bv_0^2t}}$ and by differentiation $a=-bv_0^3\sqrt{\frac{m}{(m+2bv_0^2t)^3}}$

I will firstly set out a series of hints. Try to do the questions from these alone and if you are still having difficulty then scroll down to the full solutions.

Hints

Section 2 - Physics

1 a) Work out the area enclosed by the function, taking any areas below the horizontal axis as negative.

1 b) This is a straight line therefore you need the gradient of it and any point on it.

1 c) (i) There is no need to integrate - consider what an integral calculates and use the graph.

1 c) (ii) This should be straightforward from your knowledge of velocity-time graphs.

1 d) Apply $F=ma$.

1 e) Apply $F=ma$ once more and then $P=Fv$

1 f) The graphs to be drawn are related to the one given by the area under it and the gradient of it.

2 a) Consider the force(s) acting on the stone and what their resultant can be equated to.

2 b) If the question seems unfamiliar to you then don't be put off and follow the instructions.

2 c) The initial velocity is $v_0$ and you have an expression for the velocity as a function of time. Let $v$ equal $v_0$ and work from there.

2 d) The most important aspect of both graphs is their behaviour for $t=0$ and as $t\rightarrow \infty$.

Full solutions Section 2 - Physics

1 a) Work out the area enclosed by the function, taking any areas below the horizontal axis as negative.

The areas enclosed by each triangle or rectangle, counting from left to right are as follows:

$3600$

$28800$

$4000$

$-4000$

$-28800$

$-3600$

If you add these together then you obtain $0$ so the train is back at the station.

1 b) This is a straight line therefore you need the gradient of it and any point on it.

The gradient is $\frac{-40}{200}=\frac{-1}{5}$ and a point is $(1100,0)$.

$v-0=\frac{-1}{5}(t-1100)$

$v=220-\frac{t}{5}$

1 c) (i) There is no need to integrate - consider what an integral calculates and use the graph.

The integral represents the area of the triangle $DE(1500,0)$ which is $-4000$.

1 c) (ii) This should be straightforward from your knowledge of velocity-time graphs.

This represents the distance travelled between $D$ and $E$.

1 d) Apply $F=ma$.

$a=\frac{-1}{5}$

$F=ma=2000N$

This force arises through braking and friction/resistance.

1 e) Apply $F=ma$ once more and then $P=Fv$

$F=ma=10000\times\frac{40}{180}=2222\frac{2}{9}$

$v=20$

$P=Fv=44444\frac{4}{9}W$

1 f) The graphs to be drawn are related to the one given by the area under it and the gradient of it.

2 a) Consider the force(s) acting on the stone and what their resultant can be equated to.

The only horizontal force on the stone is the drag force. It opposes the velocity; if $v$ is positive then the drag force should be negative so a minus sign is required in the equation. The resultant force is equal to $ma$ in accordance with Newton's second law.

2 b) If the question seems unfamiliar to you then don't be put off and follow the instructions.

2 b) (i) $[\frac{-1}{2v^2}]^v_{v_0}=\frac{1}{2v_0^2}-\frac{1}{2v^2}=\frac{-bt}{m}$

2 b) (ii) $\frac{1}{v^2}-\frac{1}{v_0}^2=\frac{2bt}{m}$

$\frac{1}{v^2}=\frac{2bt}{m}+\frac{1}{v_0^2}=\frac{2btv_0^2+m}{mv_0^2}$

$v^2=\frac{mv_0^2}{2btv_0^2+m}$

2 c) The initial velocity is $v_0$ and you have an expression for the velocity as a function of time. Let $v$ equal $\frac{v_0}{2}$ and work from there.

Let $v=\frac{v_0}{2}$

$\frac{v_0^2}{4}=\frac{mv_0^2}{2btv_0^2+m}$

$4m=2btv_0^2+m$

$t=\frac{3m}{2bv_0^2}$

2 d) The most important aspect of both graphs is their behaviour for $t=0$ and as $t\rightarrow \infty$.

$v=v_0\sqrt{\frac{m}{m+2bv_0^2t}}$ and by differentiation $a=-bv_0^3\sqrt{\frac{m}{(m+2bv_0^2t)^3}}$