Alan

Let taxi speed be v km per hour, so truck speed is 0.75v kph.

Truck travels for 3 hours, so distance travelled by truck in that time = 0.75v*3 = 9v/4 km

Taxi travels for 2 hours, so distance travelled by taxi in that time = 2v km

These must total 360 km so:   2v + 9v/4 = 360 or 17v/4 = 360

v = 4*360/17 = 84.71 kph

So truck speed = 0.75*4*360/17 = 63.53 kph

$$Turn $14\frac{2}{3}$ into an improper fraction = $\frac{44}{3}$\\\\
Divide this by 4 to get $\frac{11}{3}$ or $3\frac{2}{3}$$$

xvxvxv said: 

"alan .. for Q 2 The answer in the model answer is A .. so the area in part D is positive !"

No! The question asks for net displacement not total distance travelled.  When the velocity is negative the player is moving in the opposite direction to when the velocity is positive, so he is reducing his displacement.  If the model answer is A then it is wrong!

(Total distance travelled would be 20m).

Let r = length of riding course, s = length of swimming course and n = length of running course.

Then s = 0.2r and n = s/5 = 0.2r/5 so total length = r + 0.2r + 0.2r/5 = 1.24r

This equals 62km so 1.24r = 62;  r = 62/1.24 or:

r = 50 km

s = 0.2r = 10km

n = s/5 = 2km

For question 2) you need to find the area under the curve as distance is the integral of velocity with respect to time. The areas of each section are easy to find as they are comprised of triangles and rectangles.  The only point to note is that the last section will contribute a negative area.  (You should find B to be the answer)

If he originally had M then if he gave (2/5)M to his wife he would have (3/5)M left.  If he spent half of this he would have half of it left.  Half of (3/5)M is (3/10)M.  This is equal to #300, so

$$\frac{3}{10}M=#300\\M=\frac{10}{3}\times #300\\
M=#1000$$ 

If you can do C then D should be straightforward.  For E first divide through by t.

Volume = mass/density so

Volume = 75g/25g/cm³

Volume = 3 cm³

You are given the mass!!

I think you will need to solve this numerically.  The graphs below might help:

The first two solutions above are at θ = 29° and 32°

On second thoughts, a numerical solution might be overkill!  If θ - 26° is in the first quadrant, then cos(θ - 26°) = sin(90°- θ + 26°).  In this case 3θ = 116° - θ or 4θ = 116° or θ = 29°.

Melody, I tend to think of differential equations as a large subject in its own right, though you can, of course, think of it as a subset of calculus.

A danger of having too many subgroups is that some topics might fit into more than one (e.g. ODEs or Calculus?).

(Incidentally, Matrices would be part of Linear Algebra).