Displacement of a partical along a straight line at time t is x=a0+a1t+a2t2 then accleration is
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x = a^0 + at + a^2.t^2
x = 1 + a.t + a^2.t^2
v = dv/dt = d(1 + A.t + a2t2)/th
v = a + 2a^2.t
A = dv/dt = d(a + 2a^2.t)/dt
A = 2a^2
Hence, the Acceleration (A) is 2 times the square of the constant 'a'.
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x = 1 + a.t + a^2.t^2
v = dv/dt = d(1 + A.t + a2t2)/th
v = a + 2a^2.t
A = dv/dt = d(a + 2a^2.t)/dt
A = 2a^2
Hence, the Acceleration (A) is 2 times the square of the constant 'a'.
Mark my answer as brainliest !!!
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