SOLUTION: A car moving in a straight line passes through two points A and B, which are 120m apart, with speeds 72m/s and 56m/s respectively. Calculate the (i) retardation of the car (ii) dis

(a) Let "m" be the mass of the car, and let "a" be the deceleration.
    Write the mechanic energy conservation law

      maS =  - .    (1)


    Here " ma "  is the deceleration force (by the Newton's second law); 

         " S "   is the distance between the points;

         " maS " is the mechanical work performed by the force "ma";

           and    are velocities of the car at given points.


    Formula (1) says that the change of kinetic energy (right side) is equal to
    the work done by the force  ma  on the distance S (left side).


    Cancel "m" in both sides of the equation (1) and substitute given values. You will get

      a*120 =  -   56^2/2 - 72^2/2

      120a  = -1024

         a = -1024/120 = -8.533   (rounded).


      Thus deceleration is  8.533 m/s^2.   It is the ANSWER for part (a).

                                           The sign  " minus, - "  means deceleration as opposite to acceleration.

(b)  To solve (b), use the same formula (1), but this time in different form

        maS = -    (2)    (because the speed at the stop condition is zero now)


     This formula (2) expresses the same as formula (1) above.


     Cancel "m" in both sides of the equation (2) and substitute given values. You will get


        -8.533*S = -

         8.533*S = 1568

               S = 1568/8.533 = 183.76  (rounded)

     Thus the deceleration distance to get full stop is 183.76 meters. 

     It is the ANSWER to part (b).

(c)  To find time "t" to stop after passing point B, use the formula 

        S = .


     Substitute S = 183.76 m  and  a = 8.533 here and get

        t =  =  = 6.563 seconds.


     It is the ANSWER to part (c).