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[cashem'up]

this is the exact answer:
3.92 m/s
1.5 m
thankyou 

this is the exact answer:
3.92 m/s
1.5 m
thankyou 

hey halosh i got it..... i initially took g=10 lolzz.....
now we have a diagram something like this
ok now for the first part it will be easier to take the system as a whole
so lets find
the total resisting force
1.Weight is 9.8m
2. downward force due to weight is 0.6x9.8m = 5.88m
3. due to friction is F= 1/4 x 0.8(cos of arcsin0.6) x 9.8m= 1.96m
Total Resisting- 7.84m
Total Forward = 2mg =  19.6m
therefore resultant is
19.6m - 7.84m = 3ma
11.76=3a
a=3.92 m/s2

[halosh92]

hey halosh i got it..... i initially took g=10 lolzz.....
now we have a diagram something like this
ok now for the first part it will be easier to take the system as a whole
so lets find
the total resisting force
1.Weight is 9.8m
2. downward force due to weight is 0.6x9.8m = 5.88m
3. due to friction is F= 1/4 x 0.8(cos of arcsin0.6) x 9.8m= 1.96m
Total Resisting- 7.84m
Total Forward = 2mg =  19.6m
therefore resultant is
19.6m - 7.84m = 3ma
11.76=3a
a=3.92 m/s2

thx alot
sorry for the trouble
 

[cashem'up]

haha no worries man no worries

[Sue T]

jan 2007:
can ny1 solve the last question 7(f)?

 ???no1?

[halosh92]

at 2.p.m. the position vector relative to a lighthouse of ship A is (3i+j)km and its velocity is (-i + 5j)km/h
At the same time ship B has position vector (-i + 4j)km relative to the lighthouse and velocity (2i + 3j)km/h . find, after "t"h
a) the position vector of A relative to the lighthouse.
b)the position vector of B relative to the lighthouse.
c)the position vector of A relative to B.
d) find the time when A is due north of B.

can someone just solve part (d) how to come about such questions wat do they mean exactly?

[Phosu]

A particle P is moving with constant velocity (–5i + 8j) m s–1. Find
(a) the speed of P,
(b) the direction of motion of P, giving your answer as a bearing.

At time t = 0, P is at the point A with position vector (7i – 10j) m relative to a fixed origin O. When t = 3 s, the velocity of P changes and it moves with velocity (ui + vj) m s–1, where u and v are constants. After a further 4 s, it passes through O and continues to move with velocity (ui + vj) m s–1.
(c) Find the values of u and v.

(d) Find the total time taken for P to move from A to a position which is due south of A.

plz tell me how to solve part d

[halosh92]

A particle P is moving with constant velocity (–5i + 8j) m s–1. Find
(a) the speed of P,
(b) the direction of motion of P, giving your answer as a bearing.

At time t = 0, P is at the point A with position vector (7i – 10j) m relative to a fixed origin O. When t = 3 s, the velocity of P changes and it moves with velocity (ui + vj) m s–1, where u and v are constants. After a further 4 s, it passes through O and continues to move with velocity (ui + vj) m s–1.
(c) Find the values of u and v.

(d) Find the total time taken for P to move from A to a position which is due south of A.

plz tell me how to solve part d

how did u do part (c) first of all?

[Phosu]

first find the new vector position using the initial velocity, for first 3 sec
(7i-10j) +(-5+8j)3 = -8i+14j

after the boat reaches this vector position it changes speed, and travels for 4s until it reaches origin O which is O so
0=(-8i+14j)+(ui+vj)4
0=(-8+4u)i+(14+4v)j

now solve seperately
-8+4u=0  u=2
14+4v=0 v=-3.5
new velocity (2i-3.5j)

[halosh92]

at 2.p.m. the position vector relative to a lighthouse of ship A is (3i+j)km and its velocity is (-i + 5j)km/h
At the same time ship B has position vector (-i + 4j)km relative to the lighthouse and velocity (2i + 3j)km/h . find, after "t"h
a) the position vector of A relative to the lighthouse.
b)the position vector of B relative to the lighthouse.
c)the position vector of A relative to B.
d) find the time when A is due north of B.

can someone just solve part (d) how to come about such questions wat do they mean exactly?

could someone plzz solve this ?
 

[T.Q]

could someone plzz solve this ?
 

(i) component of A will be equal to the (i) component of B

[halosh92]

(i) component of A will be equal to the (i) component of B

ok but is there some kind of rule for these questions they come very often...i mean i dont rili understand when the say
"north of B" or "east of B"
shouldnt one of the components be = 0 in such cases 

[astarmathsandphysics]

There is a page on my website astarmathsandphysics.com>a level maths notes> relative positions more on relative positions

[halosh92]

There is a page on my website astarmathsandphysics.com>a level maths notes> relative positions more on relative positions

thankyou

[Uchia]

2007 january paper question 7 last part

and thanks

[astarmathsandphysics]

cie or edexcel