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Asked by arushkumar6607 | 16 Jun, 2022, 08:46: AM

Expert Answer

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

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

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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

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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

Answered by Thiyagarajan K | 16 Jun, 2022, 10:07: AM

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