Mathematics: Post your doubts here!

[kitkat <3 :P]

6 ?

for first part integrate the equation
u will get
s=2t^2 - t^4/64 now u are given with s put it in the equation or quadratic k master ho aap

 

[kitkat <3 :P]

Make the graph first. Then it will be easy. Meaning do b first and then a

so do u mean that we should caculate the total distance from graph :/?

 

[Thought blocker]

for first part integrate the equation
u will get
s=2t^2 - t^4/64 now u are given with s put it in the equation or quadratic k master ho aap

ty

 

[Thought blocker]

so do u mean that we should caculate the total distance from graph :/?

Area under graph = Distance

 

[kitkat <3 :P]

Area under graph = Distance

ye pata hai mujko .....like distance nikal k kuch karna hai UGHHHH

 

[unique111]

so do u mean that we should caculate the total distance from graph :/?

No, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.

 

[Thought blocker]

No, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.

I got it, ty.

 

[kitkat <3 :P]

No, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.

thank u

 

[Thought blocker]

No, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.

Q6 karde bhai

 

[kitkat <3 :P]

No, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.

btw we can do it by a=v-u/t too rightt??

 

[CЯeScɘnt]

Q6 bhi kardo

for part 1
use v=u+at
u=0, a and t given
find v.
v=15
now for the deceleration part V will be its u
then again
v=u +at
v=0, a= -0.0375, u=15
solve and find t
part 2)

the red lines above are for part 3,
now for 3,take 7.5/t=gradient of the lines
this will give you two answers of t.
clear?

 

[CЯeScɘnt]

Q6 karde bhai

sketching os required in part two,so cant apply this on part one....we have to use the formula.

 

[Snowysangel]

ok look for potential energy we'll use mgh
1.2*10*1.5
we calculated the tension 15 N
W=F*s
=15*1.5
=22.5
work donne by pulling force = gain in P.E -loss in K.E.

Why isn't it work done by pulling force-work dong against resistance=change in mechanical energy?

 

[kitkat <3 :P]

sketching os required in part two,so cant apply this on part one....we have to use the formula.

i think you are talking about 5 ._.

 

[CЯeScɘnt]

i think you are talking about 5 ._.

ryt.

 

[Thought blocker]

i think you are talking about 5 ._.

yup I want 6.

 

[kitkat <3 :P]

Why isn't it work done by pulling force-work dong against resistance=change in mechanical energy?

actually in the question it was written fixed smoot pulley so i assume no friction will be there

 

[Thought blocker]

ryt.

I got 5

 

[Snowysangel]

actually in the question it was written fixed smoot pulley so i assume no friction will be there

Oh sorry what question? I kinda skipped over most of these comments cause I didn't have the time

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_42.pdf
Q4ii)