Good evening Annela! I will look at this question later.

For Q7(ii), the last term in the 3rd step is expanded wrongly.

When you expand this:
... - 2 (4√3 - 3 + 4 - √3) cos 60°

Remember that “cos 60°” must be multiplied the same way as “-2”. This means that every number in the brackets will get “cos 60°”.

Your mistake is in the 4th step where only “-2√3” gets multiplied with “cos 60°” and “4√3 - 3 + 4” don’t.

Actually the last term in the 3rd step could be simplified in a very easy way: just change “cos 60°” to ½ and ½ will cancel out with the “2” in front of the brackets.

Then, you’re only left with:
... - (4√3 - 3 + 4 - √3)

Hope it helps.

For Q9, from O to X, the rope was slack so the bungee jumper would accelerate due to gravity. In Physics, all free-falling objects increase their speed at the same rate of 10m/s every second, as the gravitational acceleration is constant at about 10 m/s² (ignoring air resistance).

Upon reaching X, the rope starts to tighten and the jumper will decelerate. Point Y is the lowest point of the jump because the rope is fully stretched at that point of time.

For Q9(a)(i), a = 10. To get velocity, we integrate acceleration with respect to time. Hence,
v = 10t + c

Since the jumper falls from rest, when t=0, v=0.
0 = 10(0) + c
c = 0

Therefore,
v = 10t

Sub t=4,
v = 10(4) = 40 m/s

For Q9(ii), we integrate velocity to get distance (or displacement).

From Q9(i), v = 10t.
d = (10t²)/2 + c = 5t² + c

Similarly, when t=0, d=0.
0 = 5(0)² + c
c = 0

Hence,
d = 5t²

Sub t=4,
d = 5(4)² = 80 m

I don’t know if you take Physics, but if you do, you can sketch a speed-time graph and it’s linear because the gradient (which represents gravitational acceleration) is constant. The area under the speed-time graph (shape of a right-angled triangle) is the distance travelled, and it’s going to give you the same answer.

For Q9(iii), we integrate dV/dT to get V.

dV/dT = 10 - kT
V = 10T - (kT²)/2 + c

When T=0, the jumper was at X (just started to decelerate), so her velocity was still 40 m/s, as found in Q9(i).

Sub T=0 and V=40,
40 = 10(0) - k(0)²/2 + c
c = 40

Hence,
V = 10T - (kT²)/2 + 40

When T=3, it comes to rest at Y. Therefore, V=0.
0 = 10(3) - k(3)²/2 + 40
0 = 30 - (9/2)k + 40
9/2k = 70
k = 70 ÷ (9/2)
= (70 x 2)/9
= 140/9 (shown)

Hope you understand what I wrote.

Good evening Annela! Currently working on Q10ii.

I will write up a version of Q9 before 4 am.