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comment on you username: "NO , I procrastinate"Thank you, kind sir. Wish you the best of luck with your exams, assuming you are sitting one.
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comment on you username: "NO , I procrastinate"Thank you, kind sir. Wish you the best of luck with your exams, assuming you are sitting one.
this is how to factorise in cases like these. you can substitute this value in V to find max velocity.QP: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_43.pdf
Q7 pii
MS: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_43.pdf
Q7 pii, I know that when velocity is maximum acceleration is zero but idk how to solve this part, I am getting t=0 , when solving a=0.
Help Me.
Is this m1?The top of a cliff is 40 metres above the level of the sea. A man in a boat, close to the bottom of the cliff, is in difficulty and fires a distress signal vertically upwards from sea level.
This signal is above the level of the top of the cliff for (root) 17 seconds
Find the speed of projection of the signal.
Someone pls help thx!!!
do you understand it after video or should i explain it?View attachment 41973
plz explain iii and iv
can u explain it?do you understand it after video or should i explain it?
this is like an animtion to help you see what happens.can u explain it?
the video didnt work... and my internet connection is too slow ryt now to load it on proxy or anything... but thankss anyways for that!
I'm having the same problem.. somebody please explain thishttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Pleaseee help me with qn no. 7(i). From the marking scheme, why is K.E = 0.5 x Vapproach - Vreturn? Change in K.e is usually (final speed - initial speed), so why have they done the other way round here?
I'll be really grateful for your help
Suchal Riaz or anyone else
I'm just gonna explain the whole part..hi i really need help im this question
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_42.pdf
Q7, the second part of the part ii
thanks for the animation ...daredevil please remind me to explain that part to you. now i have to go. sorry for being busy and delaying your doubt for long time.
ohh it moves for a maximum height... until its velocity becomes zero and then falls back to the 'black line' where it comes to rest?daredevil please remind me to explain that part to you. now i have to go. sorry for being busy and delaying your doubt for long time.
First partplease explain step by step
hanksI'm just gonna explain the whole part..
For this part, we have:
u = 0 (initial velocity)
s = 0.6 m
v = ? (thats what we're looking for)
a = 5.846 (from part i)
lets use the equation v^2 = v^2 + 2as
v^2 = 0 + (2 x 5 x 0.6)
v = 2.65 m/s
For A:
There is no force acting on it, so forward force = 0
The only force is the horizontal component of the weight; mgsin(theta)
From F = ma,
0 - mgsin(alpha) = ma
-gsin(alpha) = a
a = -10 x 16/65
a = (-32/13) m/s^2
Now lets find the maximum distance travelled by A after B hits the ground (v = 0)
The final speed of B = initial speed of A
v^2 = u^2 + 2as
0 = (2.65)^2 + (2 x -32/13 x s)
s = 1.425m
So the distance of A from P will be:
2.5 - (0.6 + 1.425)
distance = 0.475m
Hope this helped
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