Mathematics: Post your doubts here!

[RoOkaYya G]

but no mass was provided

-__________-

a= -g sin alpha is the formula

no need of mass

 

[emilyy]

ur right.... i got the answer
tnx

 

[Abdel Moniem]

how to find value of alpha??

u=2.5m/s v=1.5m/s s=4m use v^2=u^2+2as v^2-u^2/2s=a=(1.5)^2-(2.5)^2/8=a=-1/2m/s^2
ii) apply newtons 2nd law sum of force in direction of motion-the sum of forces in opposite to motion=mass*acceleration
-mgsinѲ=ma
-m(10sinѲ)=-1/2m
10sinѲ=1/2
sinѲ=1/20

 

[emilyy]

i alrdy got the answers
bt tnx anyway..dats so kind of u

 

[RoOkaYya G]

ur right.... i got the answer
tnx

wlcme

 

[RoOkaYya G]

u=2.5m/s v=1.5m/s s=4m use v^2=u^2+2as v^2-u^2/2s=a=(1.5)^2-(2.5)^2/8=a=-1/2m/s^2
ii) apply newtons 2nd law sum of force in direction of motion-the sum of forces in opposite to motion=mass*acceleration
-mgsinѲ=ma
-m(10sinѲ)=-1/2m
10sinѲ=1/2
sinѲ=1/20

use a=-gsin alpha
its easier. use it for inclined planes. it always works. no need to go through all tht pain

 

[emilyy]

she is right

 

[Abdel Moniem]

actually y dont we take h as 160 itself?
we shld always take sin theta?

we calculated potential energies at a distance x along the incline not when the particle reached the top at A

 

[RoOkaYya G]

we calculated potential energies at a distance x along the incline not when the particle reached the top at A

ohh ok ty

 

[Abdel Moniem]

ohh ok ty

your welcome

 

[RoOkaYya G]

part (iii) how to find a??
i replaced but cnt find it using kinematics formulas :S
mark scheme says u= 4/3 but value of k= 4/3 :S
help!!! urgent!!! plz Abdel Moniem

 

[Abdel Moniem]

View attachment 48056

part (iii) how to find a??
i replaced but cnt find it using kinematics formulas :S
mark scheme says u= 4/3 but value of k= 4/3 :S
help!!! urgent!!! plz Abdel Moniem

Integrate ∫ (0.04t-0.0001t^2+k)dt set your limits from 400 seconds to 0 seconds
0.04t^2/2-0.0001t^3/3+kx=1600
0.04(400)^2/2-0.0001(400)^3/3+400k=1600
3200/3+400k=1600
3200+1200k=4800
k=4/3
at Maximum speed a=0m/s^2 so differentiate with respect to t
0.04-(2*0.0001t)=0
t=200 seconds
v=0.04t-0.0001t^2+k at t=200 seconds and k=4/3
v=16/3 m/s

 

[RoOkaYya G]

Integrate ∫ (0.04t-0.0001t^2+k)dt set your limits from 400 seconds to 0 seconds
0.04t^2/2-0.0001t^3/3+kx=1600
0.04(400)^2/2-0.0001(400)^3/3+400k=1600
3200/3+400k=1600
3200+1200k=4800
k=4/3
at Maximum speed a=0m/s^2 so differentiate with respect to t
0.04-(2*0.0001t)=0
t=200 seconds
v=0.04t-0.0001t^2+k at t=200 seconds and k=4/3
v=16/3 m/s

third part i said. the acceleration only. :S

 

[Thought blocker]

third part i said. the acceleration only. :S

Link bhezdo :/

 

[RoOkaYya G]

Link bhezdo :/

picture hain
link nhi hain
june 2014 p 41 quest 7

 

[Thought blocker]

picture hain
link nhi hain
june 2014 p 41 quest 7

To me kha se deku?

 

[RoOkaYya G]

To me kha se deku?

http://www.gceguide.com/search/label/AS & A Level : Mathematics (9709)
here
open the paper nw

 

[Thought blocker]

http://www.gceguide.com/search/label/AS & A Level : Mathematics (9709)
here
open the paper nw

I looked into the paper, the thing you dont get is acceleration in the iii part, right?

 

[Abdel Moniem]

Cyclist P reached B at speed 5 m/s and travelled a distance from B to C of 1400 m

third part i said. the acceleration only. :S

Cyclist P moves with constant speed of 5 m/s t=1400/5=280 seconds to travel from B to C. at t=400 second the speed of cyclist Q is 4/3m/s this represents the intial velocity. plug that into the equation below
use s=ut+1/2at^2
1400=4/3*280+0.5(280)^2a

 

[Thought blocker]

Cyclist P reached B at speed 5 m/s and travelled a distance from B to C of 1400 m

Cyclist P moves with constant speed of 5 m/s t=1400/5=280 seconds to travel from B to C. at t=400 second the speed of cyclist Q is 4/3m/s this represents the intial velocity. plug that into the equation below
use s=ut+1/2at^2
1400=4/3*280+0.5(280)^2a

How is speed of cyclist Q 4/3?