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Asked by arushkumar6607 | 16 Jun, 2022, 08:46: AM
when  time is greater than zero and less than 4 , displacement is linearly increasing

let x = a + b t  ................................(1)

at t = 0 , x = 0 , hence we get  a = 0

at t = 0 , x = 4 , hence we get b = 1

hence during the interval t=0 to t=4 , displacement x = t

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During the interval t=4 to t = 10 , displacement x remains constant as 4

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when  time is greater than 10 and less than 14 , displacement is linearly decreasing

let in this time interval ,  x = a + b t  ................................(2)

at t = 10 , x = 4 , hence we get  a + 10 t = 4  ............... (3)

at t = 14 , x = -4 , hence we get a + 14 t  = -4  .................(4)

By solving eqn.(3) and eqn.(4) , we get a = 24 and b = -2

hence during the interval t=10 to t=14 , displacement x = 24-2t
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when  time is greater than 14 and less than 18 , displacement is linearly increasing

let in this time interval ,  x = a + b t  ................................(5)

at t = 14 , x = -4 , hence we get  a + 14 t = -4  ............... (6)

at t = 18 , x = 0 , hence we get a + 18 t  = 0  .................(7)

By solving eqn.(6) and eqn.(7) , we get a = -18 and b = 1

hence during the interval t=10 to t=14 , displacement x = t-18
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hence given displacement as a function of time t is

 Displacement equation time interval x = t 0 < t < 4 x = 4 4 < t < 10 x = 24 - 2t 10 < t < 14 x = t - 18 14 < t < 18

From above expressions we get displacement at required time as given in table

 t x 1 1 2 2 4 4 6 4 11 2 13 -2 16 -2
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Velocity dx/dt  as a function of time t is

dx/dt  = 1  , when  0 < t < 4

dx/dt = 0 , when 4 < t < 10

dx/dt = - 2 , when 10 < t < 14

dx/dt =  1 , when 14 < t < 18

From above expressions we get velocity at required time as given in table

 t dx/dt 1 1 2 1 4 0 6 0 11 -2 13 -2 16 1

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since velocity is constant in all time intervals, acceleration d2x/dt2   is given as

d2x/dt2   = 0   , when 0 < t < 18

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