v-final speed

u-initial speed

a-acceleration

t=time

Formulae:v=u+at

]]>Moral of the story, If u cant handle this,don't choose A levels go poly! Trust me its not worth the tears of blood shed X_X.

]]>Next time don't forget to write down your working (even though Miss Loi knows they're all on your head!).

So to complete your answers with some explanations:

1. In a speed/time graph, the distance covered in 10 sec will be the area under the graph bounded by 0 ≤ `t` ≤ 10, which is the area of a trapezium (see shaded portion below).

Hence solve for:

(1/2)(2 + `u`)(10) = 60

Hope students still remember the formula to find the area of a trapezium!

2. Let the speed be `x` at 16 seconds. If you draw some lines on the relevant coordinates, you'll actually come face to face with two similar triangles (see the shaded portion below):

Hence solve for `x`:

(10-`x`)/`x` = 1/14

3. Hope Miss Loi's sketching skills are good enough to illustrate your 'descriptive' answers for clarity:

As mentioned in the main entry, students really, really need to understand the concepts of gradients in different graphs.

Also remember to label all points on the axes (i.e. distance) else marks might be lost!

]]>2. 9 and 1/3 m/s

3a. starting from zero, a curved line with an increasing slope, followed by a straight line with constant slope, then lastly a curved line with a decreasing slope.

3b. horizontal line at a=0.8, followed by horizontal line at a=0, lastly horizontal line at a= -2/3

hopefully my mental calculations aren't too hopelessly wrong.

]]>2nd one if i am lucky is 4 - 1/16 = 64/16 - 1/16= 63/16 = 3 15/16!

i dont know the rest ðŸ˜› how are you going to release ans?

(i'm one of the few who don't take amaths.geez..)