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Theory of Quadratics (Videos, Notes, Quiz with Correction) - Introduction

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  • April 15, 2022, 10:55 p.m.
    KENG ElSON

    Please help me solve this problem 

  • March 12, 2022, 3:05 a.m.
    Kelvino237

    Question 2 june 2018 from your calculations e=1/3

  • Feb. 22, 2022, 5:20 a.m.
    Alsojack

    Page 3

  • Feb. 22, 2022, 5:19 a.m.
    Alsojack

    Page 2

  • Feb. 22, 2022, 5:18 a.m.
    Alsojack

    Solution to June 2018 qn 7 by Johan

  • Feb. 21, 2022, 7:02 p.m.
    Mbapou rene

    This is thé solution to the question (# Q6 PMM GCE jeune 2018)

  • Feb. 21, 2022, 5:23 a.m.
    Faith

    Question 7 June 2018

        Using : P/v = 14.28×1000/6m/s

                            =2371.6N

     

     

    a- considering the car and the carriage as a single body the acceleration of the system 

       F - R = ma

    2371.6 - (639 - 280) = ( 1400 +700)a

    2021.6 = 2100a

    a= 0.96m/s^2

     

     

    b- T - 280 = 700a

    T - 280 = 490 

    T = 770N

     

     

    c- F = P/V

    = 14.28 × 1000/12

    = 1190N

     

    So, F - R = ma 

    1190-(630-230) = 1400+ 700a

    280 = 2100a

    a = 0.13 m/s^2

     

    d- time taken ,

         We have first,280=700a

                                      a=-0.4m/s^2

    Retardation of its carriage at rest 

         Using 

     V= u + at

    0= 12 + ( -0.4)t

    -12= -0.4t

    T= 30 s

     

           Faith's solution

    • Feb. 21, 2022, 5:25 p.m.
      Wanyu

      That was good Faith but you made a mistake when calculating the acceleration in the (a) part of the question. You were supposed to add the resistances not subtract them. F-R=ma, R = 639 + 280 = 919, a = (F-R)/m = (2371.6-919)/(1400+700) = 0.7 in meters per squared seconds

  • Feb. 20, 2022, 10:47 p.m.
    Mbapou rene

    This  is the solution to the question on impacts (# GCE june 2018 Q7) 

     

    • Feb. 21, 2022, 5:35 p.m.
      Wanyu

      You've not uploaded any solution Rene, please do so this evening

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