How was Mechanics Paper 42!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[talha196]

Speed is 2.6

it was asking about distance covered in T/2 seconds in the 2nd part of question 4

 

[Jigar]

6.5m

 

[talha196]

6.5m

no seriously i dont remember whether it asked for distance at T/2 or speed at T/2 .... can anyone cnfirm that ?

 

[Jigar]

no seriously i dont remember whether it asked for distance at T/2 or speed at T/2 .... can anyone cnfirm that ?

it asked for distance at t/2

 

[Your-Blood]

no seriously i dont remember whether it asked for distance at T/2 or speed at T/2 .... can anyone cnfirm that ?

They asked distance

 

[talha196]

They asked distance

so how did you calculate it i just recalculated now and i still get 3.25 m

 

[Jigar]

They asked distance

what was your bro

 

[Jigar]

t = s........therefore T/2=s/2......but yet use equation of motion and then get time.....replace t in v=u+at.....and u will be home

 

[Your-Blood]

Question No.4

m g sin(θ) = m a ( cancel m at both sides )
g sin(θ) = a

a = 16/65 x 10
a = 2.46 ms^-2


V^2 = U^2 + 2 as
8^2 = O^2 + 2(2.46)(s)
S= 13m

Time taken to travel S
13 = 1/2 (2.46) T^2

T = 3.25s


T/2 = 1.63


S when T/2

1/2 (2.46)(1.63)^2
S = 3.25m

 

[talha196]

t = s........therefore T/2=s/2......but yet use equation of motion and then get time.....replace t in v=u+at.....and u will be home

it wasnt like that
find time using v=u+at
acceleration is 10*sin@ = 10*16/65 = 2.46
8=0+ 2.46*T
T=3.252
T/2= 1.626
for finding distance apply s=ut + 0.5at^2
s=0(t) +0.5*2.46*1.626^2
s=3.25/
please correct me if i am wrong

 

[Your-Blood]

But I think the question was to calculate time when S was S/2.

I seriously can't remember what was the question

 

[talha196]

Question No.4

m g sin(θ) = m a ( cancel m at both sides )
g sin(θ) = a

a = 16/65 x 10
a = 2.46 ms^-2


V^2 = U^2 + 2 as
8^2 = O^2 + 2(2.46)(s)
S= 13m

Time taken to travel S
13 = 1/2 (2.46) T^2

T = 3.25s


T/2 = 1.63


S when T/2

1/2 (2.46)(1.63)^2
S = 3.25m

hey in frst part it also required the speed at S/2 rite????

 

[Your-Blood]

it wasnt like that
find time using v=u+at
acceleration is 10*sin@ = 10*16/65 = 2.46
8=0+ 2.46*T
T=3.252
T/2= 1.626
for finding distance apply s=ut + 0.5at^2
s=0(t) +0.5*2.46*1.626^2
s=3.25/
please correct me if i am wrong

This is exactly I'm doing right now.
But I think question was something else :/

 

[talha196]

i

But I think the question was to calculate time when S was S/2.

I seriously can't remember what was the question

as far as i remember it asked for the speed at S/2

 

[Your-Blood]

i
as far as i remember it asked for the speed at S/2

Yeah but that was in first part I guess. :/

 

[talha196]

This is exactly I'm doing right now.
But I think question was something else :/

i dont think so i am also confused about the statement of the question

 

[raiyanfarrukh]

bro for c the initial time was 20 and the initial dist. was 92!
i did that way and got the correct ans!

same thing i took total distance as 117m and final time!!

 

[raiyanfarrukh]

nobodys helping me around q4 last part :'( :'(

hey i even don't remember what type of question was that and everything in the paper was easy except q6 and q711 so don't worry!!why is everyone asking about q4??

 

[raiyanfarrukh]

Question No.4

m g sin(θ) = m a ( cancel m at both sides )
g sin(θ) = a

a = 16/65 x 10
a = 2.46 ms^-2


V^2 = U^2 + 2 as
8^2 = O^2 + 2(2.46)(s)
S= 13m

Time taken to travel S
13 = 1/2 (2.46) T^2

T = 3.25s


T/2 = 1.63


S when T/2

1/2 (2.46)(1.63)^2
S = 3.25m

i thinks that's what i got!!

 

[talha196]

i thinks that's what i got!!

brother do you remember the question statement?