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    Motion in Straight Line
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    A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

    As the length of one step of the drunkard is 1 m long and takes one second to take one step, therefore the speed of drunkard is 1m/s. When he walk 5 steps forward, its position is 5m at time 5s, then he walks 3 steps backward, therefore the position of drunkard at 8 sec. becomes 2m

    space space space space space space space therefore                 At       t = 5s             x = 5m
    and                         at       t = 8s             x = 2m
                                  at       t = 13s           x = 7m
                                  at       t = 16s           x = 9m
                                  at       t = 24s           x = 6m
                                  at       t = 29s           x = 11m
    and so on

    The position time graph is as shown in figure.
    As the length of one step of the drunkard is 1 m long and takes one s
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    A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

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    A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?
    Speed space of space jet space airplane comma space straight v subscript jet space equals space 500 space km divided by hr space relative space spedd space of space product space of space combustion space straight w. straight r. to space the space plane space is comma space straight v subscript smoke space equals space minus 1500 space km divided by hr space Speed space of space the space product space of space combustion space straight w. straight r. to space ground space equals space straight v apostrophe subscript smoke Relative space speed space of space space products space of space combustion space with space respect space to space the space airplane comma space space space space space straight v subscript smoke space equals space straight v prime subscript smoke space – space straight v subscript jet space end subscript space space space – space 1500 space equals space straight v prime subscript smoke space end subscript – space 500 space straight v prime subscript smoke space end subscript space space space space space space space space space space space space space space space space space equals space – space 1000 space km divided by straight h space

    Negative sign implies that the direction of its products of combustion is opposite to the direction of motion of the jet airplane. 
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    A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop.

    Initial space velocity space of space the space car comma space straight u equals 126 space km divided by hr space equals space 35 space straight m divided by straight s Final space velocity space of space car comma space straight v space equals space 0 space Before space coming space to space rest comma space Distance space covered space by space the space car comma space straight s space equals space 200 space straight m space Let comma space retardation space produced space in space the space car space equals space straight a Now comma space using space the space third space equation space of space motion comma space space space space space straight v squared space minus space straight u squared space equals space 2 as space rightwards double arrow space left parenthesis 0 right parenthesis squared space minus space left parenthesis 35 right parenthesis squared space equals space 2 space cross times space straight a space cross times space 200 rightwards double arrow space straight a space equals space minus 35 cross times 35 over 2 cross times 200 space equals space minus 3.06 space straight m divided by straight s squared space Time space taken space by space the space car space to space stop space is space given space by space the space first space equation space of space motion. space That space is comma space space space space space space space space straight v space equals space straight u space plus space at rightwards double arrow space space space straight t space equals space fraction numerator left parenthesis straight v minus straight u right parenthesis over denominator straight a end fraction space equals space fraction numerator left parenthesis negative 35 right parenthesis over denominator left parenthesis negative 3.06 right parenthesis end fraction space equals space 11.44 space sec
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    A car moving along a straight highway with speed of 126 km/hr is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

    Here,

    Initial velocity, u = 126 km/hr = 35m/s
    Final velocity, v = 0
    Distance travelled, s = 200m 

    Now using the equation of motion,

                          v2-u2 = 2as

    Substituting the value of v, u and s we get, 

                 (0)2-(35)2 = 2a x 200 

                           a = -3.0625 m/s2 

    Time taken by the car to stop is given by, 

               

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