Additional Mathematics ; post your doubts here!

[ThatReallyReallyWeirdDude]

Title says it all
If it's a past paper question, kindly link to it
Good luck everyone. ​

 

[Besterlinks]

Hi...

0606_w04_qp_2 > Question 12 EITHER part (ii).
(IGCSE Additional Mathematics > Winter 2004 > Paper 2 > Question 12 EITHER > Part (ii)

Can you please help me solve it?

Thanks

 

[ThatReallyReallyWeirdDude]

Hi...

0606_w04_qp_2 > Question 12 EITHER part (ii).
(IGCSE Additional Mathematics > Winter 2004 > Paper 2 > Question 12 EITHER > Part (ii)

Can you please help me solve it?

Thanks

Sorry for late reply, had my physics exam
Alright then, the particle comes at rest at t=5, so the summary of it's movement is as shown:

So the total distance travelled is the distance travelled between t=0 and t=5 (i.e. s5-s0) plus the distance travelled between t=10 and t=5 (s10-s5)
{v=1.4t-0.3t2+c
when t=0, v=0.5
v=1.4(0)-0.3(0)+c=0.5
therefore c=0.5
and v=1.4t-0.3t2+0.5} Though you woulda calculated this in (i)
So as for distance, s= integration of v
=0.7t2-0.1t3+0.5t + c
Since s=0 when t=0, c=0.
So s=0.7t2-0.1t3+0.5t
Now all you have to do is find |s5-s0| + |s10-s5|
=|7.5-0|+|-25-7.5|
=7.5+32.5=40 meters
The alternative way to understand it (instead of s5-s0 + s10-s5) is to think that first it gets to t=5, which is distance of 7.5m, then it goes back to 0 (another distance of 7.5m) and to t=10 (a distance of 25m FROM O)
So the total distance will be 2x7.5 + 25=40 meters.
Hope I helped!

 

[hredoymohammad]

0606/21/M/J/11
9 A coastguard station receives a distress call from a ship which is travelling at 15 km h–1 on a
bearing of 150°. A lifeboat leaves the coastguard station at 15 00 hours; at this time the ship is at a
distance of 30 km on a bearing of 270°. The lifeboat travels in a straight line at constant speed and
reaches the ship at 15 40 hours.
(i) Find the speed of the lifeboat.

i have solved many vector probs...but i m not getting dis question....n the picture given in mark scheme is not understandable...can sumone help

 

[ThatReallyReallyWeirdDude]

0606/21/M/J/11
9 A coastguard station receives a distress call from a ship which is travelling at 15 km h–1 on a
bearing of 150°. A lifeboat leaves the coastguard station at 15 00 hours; at this time the ship is at a
distance of 30 km on a bearing of 270°. The lifeboat travels in a straight line at constant speed and
reaches the ship at 15 40 hours.
(i) Find the speed of the lifeboat.

i have solved many vector probs...but i m not getting dis question....n the picture given in mark scheme is not understandable...can sumone help

Can you link to the past paper or tell me which paper it is?

 

[hredoymohammad]

i wrote in the first line
0606/21/M/J/11 .........that is may june 2011 paper 21 number 9

 

[ThatReallyReallyWeirdDude]

i wrote in the first line
0606/21/M/J/11 .........that is may june 2011 paper 21 number 9

Oh sorry! I missed it. :| I found it myself afterwards anyway
I'll make a diagram and be back at you!

 

[hredoymohammad]

THANX...pls have a luk at the markscheme's diagram...i dont understand it...it should have been a totally opposite digram ? :S right?

 

[ThatReallyReallyWeirdDude]

I'm so sorry I took so long!
This is the answer using trigonometry. Do you want the method using vectors?

 

[ThatReallyReallyWeirdDude]

Note: It's supposed to say we can use cosine theorem to find the distance CI* not SI* sorry!

 

[hredoymohammad]

0606/1/M/J/04 question seven
To a cyclist travelling due south on a straight horizontal road at 7 ms
01, the wind appears to be
blowing from the north-east. Given that the wind has a constant speed of 12 ms
01, find the direction
from which the wind is blowing.
i have understood upto the point in markscheme where it says 24.4 degree....but i could not understand how it led to the final answer of 020.6 ....

 

[ThatReallyReallyWeirdDude]

0606/1/M/J/04 question seven
To a cyclist travelling due south on a straight horizontal road at 7 ms
01, the wind appears to be
blowing from the north-east. Given that the wind has a constant speed of 12 ms
01, find the direction
from which the wind is blowing.
i have understood upto the point in markscheme where it says 24.4 degree....but i could not understand how it led to the final answer of 020.6 ....

Okay
You have to understand and draw your own diagram FIRST thing:

 

[hredoymohammad]

but dont u think ur wind direction is wrong? it is coming FROM north east...to SOUTH west....so arrow should be pointing downwards...n could u tell me where have u drawn dis image..so i could draw n show wat i meant

 

[ThatReallyReallyWeirdDude]

but dont u think ur wind direction is wrong? it is coming FROM north east...to SOUTH west....so arrow should be pointing downwards...n could u tell me where have u drawn dis image..so i could draw n show wat i meant

True. The only difference is that the arrow of the blue line is in the opposite direction, otherwise it's still the same.

 

[hredoymohammad]

but why are we finding dat angle u labelled ?.. y not the right one up??? n could u kindly tell me where have u drawn dis .? thx

 

[ThatReallyReallyWeirdDude]

I used photoshop.
The question asks the actual direction of the wind...the wind is the second vector...
Why would you find the one at the top? Pointless.

 

[hredoymohammad]

accha thx

 

[hredoymohammad]

brother. i have a prob in the first post of dis page... Now all you have to do is find |s5-s0| + |s10-s5|
=|7.5-0|+|-25-7.5|
=7.5+32.5=40 meters how dis is done? i have understood the alternate way... n brother do u estimate in taking values of t....i mean u randomly took t=5s..? y not others

 

[ThatReallyReallyWeirdDude]

brother. i have a prob in the first post of dis page... Now all you have to do is find |s5-s0| + |s10-s5|
=|7.5-0|+|-25-7.5|
=7.5+32.5=40 meters how dis is done? i have understood the alternate way... n brother do u estimate in taking values of t....i mean u randomly took t=5s..? y not others

No, the first part of the question shows that the particle has instantaneous rest at t=5s, i.e. it turns around. So you have to get the distance between t=0 and t=5, and the distance travelled AFTER it turns around i.e. the distance between t=10 and t=5

 

[hredoymohammad]

oh ..at t=5, speed is 0m/s....so after dat we put values of t and found dat distance is decreasing...that is it is returning to the starting point?...is it not the way to think...correct me if i m wrong..