Q)A CAR IS TRAVELLING ON A STRAIGHT ROAD.THE MAX VELOCITY THE CAR CAN ATTAIJN IS 24 M/S.THE MAX ACCELERATION AND RETARDATION IT CAN ATTAIN IS 1 AND 4 RESPECTIVELY.THE SHORTEST TIME THE CAR TAKES FROM REST TO REST IN A DISTANCE OF 200M IS....ANS)22.4S

Dear Koka Ram ,

 Let us assume that upto distance x it accelerates and so 200-x it deaccelerates.and let the peak velocity be v.

   V = 1*t1    -->  t1 = v

  X = (1/2) * 1* t1^2  

  x = v^2/2        ----- (1)

  now for deaccleration ,

  0 - V = -4*t2  -----> t2 = v/4

  200 -x = v * v/4 - (1/2)*4*(v/4)^2     ------(2)

add (1) and (2)

200 = v² /2 + v²  / 4 - v² /8

200 = 5v² /8

v = 17.88

therefore total time = v +v/4 =  22.4 s

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