Mathematics: Post your doubts here!

[Laibaaa]

How do we calculate V for the first part?
Ive uploaded mark scheme as well

i saw the mark scheme but it was pretty confusing can anyone please tell me how to form an equation as well?

 

[DaniyalK]

How do we calculate V for the first part?
Ive uploaded mark scheme as well

i saw the mark scheme but it was pretty confusing can anyone please tell me how to form an equation as well?

Driving force = P/V
P/V - Fr = ma
30000/Vinitial - 12500(30/500) - 1000 = 1250(4)
and
30000/Vfinal -12500(30/500) - 1000 = 1250(0.2)

 

[Laibaaa]

Driving force = P/V
P/V - Fr = ma
30000/Vinitial - 12500(30/500) - 1000 = 1250(4)
and
30000/Vfinal -12500(30/500) - 1000 = 1250(0.2)

Whats - 12500 (30/500) ?

 

[DaniyalK]

Whats - 12500 (30/500) ?

component of weight down the slope.

 

[BhaiArshad]

For

Variable Acceleration

 

[<><> Ice <><>]

Made this thread with alot of helpful last minute resources for mechanics 1
https://www.xtremepapers.com/community/threads/mechanics-i-last-minute-revision-notes.41263/

 

[nehaoscar]

Part ii please!
TheJDOG

 

[Laibaaa]

component of weight down the slope.

I dont understand, how did you get that? :s

 

[farhan141]

q6 explainView attachment 53498

 

[DaniyalK]

I dont understand, how did you get that? :s

component is mgsin(theta)
mg is 12500
For sintheta, consider the triangle in the question. Length of the slope(500) is hypotenuse and the height(30) is the perpendicular. sin(theta) = 30/500
so the component is 12500(30/500)

 

[Laibaaa]

component is mgsin(theta)
mg is 12500
For sintheta, consider the triangle in the question. Length of the slope(500) is hypotenuse and the height(30) is the perpendicular. sin(theta) = 30/500
so the component is 12500(30/500)

Ohhh, got it! thankyou so much!

 

[Ahmed Aqdam]

View attachment 53509
Part ii please!
TheJDOG

Velocity after A is 10t - 0.15t^2 +C (Integrated acceleration)
This will be equal to 5 as at t=0 P is at A and that instant its velocity is 5ms-1.
So velocity after A is 10t - 0.5t^2 +5
Integrate this to get distance
5t^2 - 0.05t^3 + 5t
Time taken from O to A was 0.5s so after A will be 3.0s.
Put 2.5 in the equation to get 42.97 m after A.
Add 1.25 m from O to A. So total distance is 44.2 m.

 

[Math Buddy]

View attachment 53508 View attachment 53510

jazak Allah khair, thank you

 

[farhan141]

jazak Allah khair, thank you

Pray for me

And ignore the top line of 2nd page

 

[DeadlYxDemon]

Need a quick reply!!!

 

[Math Buddy]

Pray for me

And ignore the top line of 2nd page

Insha Allah, pray for me too!

 

[tiki-taka]

can someone explain this q..

 

[DESTROYER1198]

guys W14 p43 Q6 iii i feel that the mark scheme is wrong and if not can someone explain it for me please.

 

[hizirofaki]

thx again guys
an d good luck tommorow inshallah t eill be good

 

[DeadlYxDemon]

Hope the paper turns out to be easy or at least we could attempt 75 % of it.. :/